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Freely Falling Objects - Problems - 1

1

Height of a Cliff – Stone Dropped from Rest

Question: (I) A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.55 s. How high is the cliff?

Given

  • Initial velocity, u = 0 m/s
  • Time, t = 3.55 s
  • Acceleration due to gravity, g = 9.80 m/s²
  • Height of the cliff, h = ?

Method 1: Using the equation of motion

The stone is dropped from rest, so its initial velocity is zero. Taking downward as positive, we use:

s = ut + ½gt²

Substitute the given values:

h = 0(3.55) + ½(9.80)(3.55)²

h = 4.90 × 12.6025

h = 61.75225 m

Rounding to three significant figures:

h ≈ 61.8 m

Method 2: Using the final velocity

First, calculate the velocity just before the stone hits the ground:

v = u + gt

v = 0 + 9.80(3.55)

v = 34.79 m/s

Now use the equation that relates velocity, acceleration, and displacement:

v² = u² + 2gh

(34.79)² = 0² + 2(9.80)h

h = (34.79)² / 19.6

h ≈ 61.8 m

Final Answer: The cliff is approximately 61.8 m high.

Key idea: When an object is dropped from rest and air resistance is neglected, its displacement is proportional to the square of the falling time: h = ½gt².

2

King Kong Falling from the Empire State Building

Question (I): Estimate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before “landing.”

Given:

  • Height of the building, h = 380 m
  • Initial velocity, u = 0 m/s
  • Acceleration due to gravity, g = 9.8 m/s2
  • Air resistance is neglected.

(a) Time taken to fall

Since King Kong falls from rest, the initial velocity is zero. Taking downward as the positive direction, use the equation of motion:

s = ut + ½gt²

Substitute s = 380 m and u = 0:

380 = 0 × t + ½ × 9.8 × t²

380 = 4.9t²

t² = 380 / 4.9

t² = 77.551

t ≈ 8.81 s

(b) Velocity just before landing

Use the equation of motion that relates initial velocity, acceleration, time, and final velocity:

v = u + gt

Substituting u = 0 m/s, g = 9.8 m/s², and t = 8.81 s:

v = 0 + 9.8 × 8.81

v ≈ 86.3 m/s downward

Alternative method for part (b)

We can also calculate the final velocity directly without first calculating the time. Use:

v² = u² + 2gs

Substituting the known values:

v² = 0² + 2 × 9.8 × 380

v² = 7448

v = √7448

v ≈ 86.3 m/s downward

Final answers

(a) Time taken ≈ 8.81 s

(b) Velocity just before landing ≈ 86.3 m/s downward

Key idea

For an object dropped from rest, use s = ½gt² to find the time and v² = 2gs to find its final speed directly. The result is an ideal estimate because air resistance is neglected.

3

Ball Thrown Vertically Upward: Initial Speed and Maximum Height

Question (II): A ball player catches a ball 3.4 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

Given

  • Total time of flight: T = 3.4 s
  • Acceleration due to gravity: g = 9.8 m/s²
  • Final position is the same as the initial position.
  • Air resistance is neglected.

Part (a): Initial speed of the ball

Since the ball returns to the player’s hand, the time taken to rise to the maximum height equals the time taken to fall back to the initial position.

Time of ascent = Time of descent

t = T/2 = 3.4/2 = 1.7 s

At the highest point, the instantaneous vertical velocity is zero. Using the equation of motion,

v = u + at

At maximum height, v = 0 and a = −g.

0 = u − gt

u = gt

u = 9.8 × 1.7 = 16.66 m/s

Initial speed = 16.7 m/s

Part (b): Maximum height reached

At the highest point, the final vertical velocity is zero. We can use the equation of motion that relates velocity, acceleration and displacement:

v² = u² + 2as

Here, v = 0, a = −g and s = H.

0 = u² − 2gH

H = u²/(2g)

H = (16.66)²/(2 × 9.8)

H ≈ 14.16 m

Alternatively, using the initial speed and the time of ascent, we can calculate the height directly:

H = ut − ½gt²

H = (16.66 × 1.7) − ½ × 9.8 × (1.7)²

H ≈ 14.16 m

Maximum height = 14.2 m

Final Answers

(a) Initial speed = 16.7 m/s upward.

(b) Maximum height = 14.2 m above the player’s hand.

Key Idea

When a ball returns to its launch height under gravity, its total flight time is twice its time of ascent. At the maximum height, its velocity is zero, but its acceleration remains directed downward with magnitude g.

4

Baseball Hit Vertically Upward: Maximum Height and Time of Flight

(II) A baseball is hit almost straight up into the air with a speed of 25 m/s. Estimate (a) how high it goes, (b) how long it is in the air, (c) What factors make this an estimate?

Given

  • Initial speed: u = 25 m/s
  • Acceleration due to gravity: g = 9.8 m/s²
  • Assume the ball is caught or returns to approximately the same height from which it was hit.
  • For the initial calculation, neglect air resistance.

Part (a): Estimate the maximum height

As the baseball moves upward, gravity acts downward and reduces its upward velocity. At the maximum height, its instantaneous vertical velocity becomes zero.

Method 1: Using the equation of motion

v² = u² + 2as

At maximum height: v = 0, a = −g, s = H

0 = u² − 2gH

H = u²/(2g)

H = 25²/(2 × 9.8)

H ≈ 31.9 m

Method 2: Using the time to reach maximum height

First, use the velocity equation to calculate the time required to reach the highest point.

v = u − gt

0 = 25 − 9.8t

t = 25/9.8 ≈ 2.55 s

Now use the displacement equation during the upward motion.

H = ut − ½gt²

H = 25(2.55) − ½(9.8)(2.55)²

H ≈ 31.9 m

Estimated maximum height ≈ 32 m

Part (b): Estimate the time in the air

If the baseball returns to approximately its launch height and air resistance is neglected, the time of descent equals the time of ascent.

Time of flight = 2 × Time of ascent

T = 2u/g

T = 2 × 25/9.8

T ≈ 5.10 s

Alternative method: Using displacement

Take the launch point as y = 0 and upward as the positive direction. When the ball returns to its launch height, its displacement is zero.

y = ut − ½gt²

0 = 25t − 4.9t²

t(25 − 4.9t) = 0

t = 0 or t = 25/4.9

T ≈ 5.10 s

Estimated time in the air ≈ 5.1 s

Part (c): Why are these estimates?

The calculations above use an idealized model of the baseball’s motion. The actual height and time in the air may differ because:

  • Air resistance: Drag opposes the ball’s motion and changes its acceleration. It generally reduces the height reached compared with the ideal no-drag calculation.
  • Initial direction: The ball is hit “almost straight up,” not necessarily perfectly vertically. A small horizontal component of velocity may be present.
  • Initial speed: The stated speed of 25 m/s is an approximate value, so the calculated height and time are also approximate.
  • Return height: The 5.1 s estimate assumes the ball returns to the same height from which it was hit. If it is caught at a different height, the total time changes.
  • Gravity: The calculation treats g as constant at 9.8 m/s², which is an excellent approximation for this short flight.

Final Answers

(a) Maximum height ≈ 31.9 m, or about 32 m.

(b) Time in the air ≈ 5.10 s, provided it returns to its launch height.

(c) Air resistance, imperfect vertical motion, approximate initial speed, and the return height affect the actual results.

Key Idea

For vertical motion under gravity, the maximum height is found using v² = u² − 2gH with v = 0 at the top. If the ball returns to its original height and air resistance is neglected, the total flight time is T = 2u/g.

5

Kangaroo Jump: Time of Flight

Question (II): A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

Given

  • Maximum height: H = 1.45 m
  • Acceleration due to gravity: g = 9.8 m/s²
  • Initial and final positions are at the same height.
  • Air resistance is neglected.

Method 1: Using the maximum-height equation

At the highest point, the kangaroo’s vertical velocity becomes zero. Using the equation of motion,

v² = u² + 2as

0 = u² − 2gH

u² = 2gH

u = √(2gH)

u = √(2 × 9.8 × 1.45)

u ≈ 5.33 m/s

Now use the velocity equation to find the time taken to reach the maximum height.

v = u − gt

0 = 5.33 − 9.8t

t = 5.33/9.8

t ≈ 0.544 s

Since the kangaroo returns to its starting height, the time of descent equals the time of ascent.

Total time = 2 × Time of ascent

T = 2 × 0.544

T ≈ 1.09 s

Method 2: Direct formula for time of flight

For an object that moves vertically upward and returns to its original height, the maximum height is related to the total time of flight by:

H = gT²/8

T² = 8H/g

T = √(8H/g)

T = √(8 × 1.45/9.8)

T ≈ 1.09 s

Final Answer

Time in the air ≈ 1.09 s

Key Idea

For a vertical jump that ends at the starting height, the total time of flight is twice the time taken to reach maximum height. The result assumes constant gravitational acceleration and negligible air resistance.

6

Vertical Leap of Basketball Players: Launch Speed and Time of Flight

Question (II): The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 120 cm. (a) What is their initial “launch” speed off the ground? (b) How long are they in the air?

Given

  • Maximum vertical leap: H = 120 cm = 1.20 m
  • Acceleration due to gravity: g = 9.8 m/s²
  • Initial vertical velocity: u = ?
  • Assume the player lands at the same height from which they jumped.

Part (a): Initial launch speed

As the player jumps upward, gravity acts downward and reduces the upward velocity. At the highest point, the vertical velocity becomes zero.

Method 1: Using the equation of motion

v² = u² + 2as

At maximum height: v = 0, a = −g, s = H

0 = u² − 2gH

u² = 2gH

u = √(2gH)

u = √(2 × 9.8 × 1.20)

u ≈ 4.85 m/s

Method 2: Using the time to reach maximum height

First, calculate the time taken to reach the highest point using v = u − gt. The time can also be found from the displacement equation.

H = ½gt₁²

t₁ = √(2H/g)

t₁ = √(2 × 1.20/9.8)

t₁ ≈ 0.495 s

u = gt₁

u = 9.8 × 0.495

u ≈ 4.85 m/s

Initial launch speed ≈ 4.85 m/s upward

Part (b): Time in the air

Assuming the player lands at the same height from which they jumped and air resistance is negligible, the time of descent equals the time of ascent.

Method 1: Using symmetry of vertical motion

T = 2t₁

T = 2 × 0.495

T ≈ 0.990 s

Method 2: Using the launch speed

For a vertical jump that returns to the launch height, the total time of flight is T = 2u/g.

T = 2u/g

T = (2 × 4.85)/9.8

T ≈ 0.990 s

Method 3: Directly from maximum height

Since H = gT²/8 for a jump that returns to its original height,

T = √(8H/g)

T = √(8 × 1.20/9.8)

T ≈ 0.990 s

Final Answers

(a) Initial launch speed ≈ 4.85 m/s.

(b) Time in the air ≈ 0.99 s.

Key Idea

At the maximum height, vertical velocity is zero, but acceleration is still g downward. For a vertical jump returning to the same height, the total time in the air is twice the time taken to rise to the maximum height.

7

Speed and Distance-Time Graphs for an Object in Free Fall

Question:

(II) An object starts from rest and falls under the influence of gravity. Draw graphs of (a) its speed and (b) the distance it has fallen, as a function of time from t = 0 to t = 5.00 s. Ignore air resistance.

Given

  • Initial speed: u = 0 m/s
  • Acceleration due to gravity: g = 9.80 m/s²
  • Time interval: 0 ≤ t ≤ 5.00 s
  • Air resistance is neglected.

(a) Speed-Time Graph

Since the object starts from rest and falls freely, its speed increases uniformly because the acceleration due to gravity is constant.

Method 1: Using the equation of motion

v = u + gt

Since u = 0:

v = gt = 9.80t

At t = 0 s:

v = 0 m/s

At t = 5.00 s:

v = 9.80 × 5.00 = 49.0 m/s

Therefore, the speed-time graph is a straight line passing through the origin and reaching 49.0 m/s at 5.00 s.

Speed-Time Graph

0 5 49 t (s) v (m/s) v = 9.80t

The straight-line slope represents the constant acceleration, 9.80 m/s².

(b) Distance-Fallen-Time Graph

Method 1: Using the displacement equation

For constant acceleration, the displacement after time t is

s = ut + ½gt²

Since u = 0:

s = ½gt² = 4.90t²

Calculate the distance fallen at selected times:

Time, t (s) Distance, s = 4.90t² (m)
0 0
1 4.90
2 19.6
3 44.1
4 78.4
5 122.5

At t = 5.00 s, the distance fallen is

s = 4.90 × 5.00² = 122.5 m

Since distance is proportional to the square of time, the distance-time graph is a curve that becomes progressively steeper. It starts at the origin and bends upward.

Distance-Fallen-Time Graph

0 5 122.5 t (s) s (m) s = 4.90t²

The slope of the distance-time graph increases with time because the falling object moves faster and faster.

Method 2: Understanding the graphs from acceleration

The acceleration is constant and equal to g. Therefore, speed increases by equal amounts in equal time intervals. This produces a straight-line speed-time graph.

The distance travelled during each successive second becomes greater because the object is moving faster. Thus, distance does not increase linearly with time; it follows the quadratic relationship s ∝ t², producing a curved graph.

Final results

  • Speed-time graph: a straight line from (0 s, 0 m/s) to (5.00 s, 49.0 m/s).
  • Distance-time graph: an upward-curving parabola from (0 s, 0 m) to (5.00 s, 122.5 m).
Key idea: In free fall from rest, v = gt and s = ½gt². The speed-time graph is linear, while the distance-time graph is parabolic.
8

Speed of a Stone at a Given Height and the Two Possible Times

Question:

(II) A stone is thrown vertically upward with a speed of 24.0 m/s.

(a) How fast is it moving when it is at a height of 13.0 m?

(b) How much time is required to reach this height?

(c) Why are there two answers to part (b)?

Given

  • Initial velocity: u = +24.0 m/s
  • Final height: s = +13.0 m
  • Acceleration due to gravity: a = −9.80 m/s²
  • Initial position: s₀ = 0 m
  • Air resistance is neglected.
  • Upward direction is taken as positive.

Important Concept

When a stone is thrown vertically upward, gravity acts downward throughout its motion. Therefore, the stone slows down while rising, momentarily reaches zero velocity at its maximum height, and then speeds up while falling.

At a height below the maximum height, the stone passes the same position twice: once while moving upward and once while moving downward.

(a) Speed of the Stone at a Height of 13.0 m

We need to calculate the speed of the stone when its vertical displacement from the point of projection is 13.0 m.

Since the initial velocity, acceleration, and displacement are known, we can use the third equation of motion without first calculating time.

Method 1: Using the Third Equation of Motion

v² = u² + 2as

Substituting the given values:

v² = (24.0)² + 2(−9.80)(13.0)

Squaring the initial velocity:

v² = 576 − 254.8

v² = 321.2

Taking the square root:

v = ±√321.2

v = ±17.9 m/s

The positive and negative signs represent two different directions of motion at the same height.

While Rising

v = +17.9 m/s

Positive velocity indicates upward motion.

While Falling

v = −17.9 m/s

Negative velocity indicates downward motion.

Speed Versus Velocity

Velocity includes both magnitude and direction, whereas speed is the magnitude of velocity.

Therefore, the speed is positive for both upward and downward motion.

Speed = |v| = 17.9 m/s

Answer for (a)

The speed of the stone at a height of 13.0 m is

17.9 m/s

(b) Time Required to Reach a Height of 13.0 m

The stone reaches the specified height twice. We can calculate both times using the first equation of motion.

Method 1: Using the First Equation of Motion

v = u + at

Rearranging the equation:

t = (v − u)/a

From part (a), the two possible velocities are +17.9 m/s and −17.9 m/s.

Case 1: Time While the Stone Is Rising

During upward motion, the velocity is positive:

v = +17.9 m/s

Using the first equation of motion:

v = u + at

17.9 = 24.0 − 9.80t

9.80t = 24.0 − 17.9

t = 6.1/9.80

t₁ ≈ 0.622 s

The stone reaches 13.0 m for the first time approximately 0.622 seconds after being thrown.

Case 2: Time While the Stone Is Falling

During downward motion, the velocity is negative:

v = −17.9 m/s

Applying the first equation of motion:

v = u + at

−17.9 = 24.0 − 9.80t

9.80t = 24.0 + 17.9

t = 41.9/9.80

t₂ ≈ 4.28 s

The stone passes through 13.0 m again approximately 4.28 seconds after being thrown, this time while falling.

Method 2: Using the Displacement Equation

We can also calculate both times directly using the second equation of motion. This method does not require the velocity calculated in part (a).

s = ut + ½at²

Substituting s = 13.0 m, u = 24.0 m/s, and a = −9.80 m/s²:

13.0 = 24.0t − 4.90t²

Rearranging:

4.90t² − 24.0t + 13.0 = 0

This is a quadratic equation of the form At² + Bt + C = 0. Using the quadratic formula:

t = [−B ± √(B² − 4AC)]/(2A)

Here:

A = 4.90, B = −24.0, C = 13.0

Substituting:

t = [24.0 ± √(576 − 254.8)]/9.80

t = [24.0 ± √321.2]/9.80

t = [24.0 ± 17.9]/9.80

t₁ = (24.0 − 17.9)/9.80

t₁ ≈ 0.622 s

t₂ = (24.0 + 17.9)/9.80

t₂ ≈ 4.28 s

Both methods give the same two times. The first equation of motion uses the two velocities at the specified height, whereas the displacement equation gives both times directly.

Answer for (b)

  • Time while rising: t₁ ≈ 0.622 s
  • Time while falling: t₂ ≈ 4.28 s

(c) Why Are There Two Answers for Time?

The stone first moves upward, slows down because of gravity, reaches its maximum height, and then falls back downward. Consequently, it passes through the same height twice.

Vertical Motion Diagram

Point of projection Height = 13.0 m Rising Falling Maximum height t₁ ≈ 0.622 s t₂ ≈ 4.28 s

The stone crosses the 13.0 m level on the way up and crosses it again on the way down. The two crossings correspond to two different times.

Checking the Maximum Height

We can verify that the stone rises above 13.0 m by calculating its maximum height. At the highest point, the instantaneous velocity is zero.

Using the Third Equation of Motion

v² = u² + 2as

At maximum height, v = 0:

0 = (24.0)² − 2(9.80)H

H = (24.0)²/[2(9.80)]

H ≈ 29.4 m

Since the maximum height is approximately 29.4 m, the stone reaches 13.0 m on both its upward and downward journeys.

Understanding the Two Time Values

Quantity On the Way Up On the Way Down
Height 13.0 m 13.0 m
Velocity +17.9 m/s −17.9 m/s
Speed 17.9 m/s 17.9 m/s
Time 0.622 s 4.28 s

Final Results

  • (a) Speed at 13.0 m: 17.9 m/s.
  • (b) Time while rising: 0.622 s.
  • (b) Time while falling: 4.28 s.
  • (c) Reason for two answers: The stone passes through the same height twice, once while rising and once while falling.
  • Maximum height: 29.4 m above the point of projection.

Key Concepts

  • Use v² = u² + 2as when velocity, acceleration, and displacement are involved, but time is not required.
  • Use v = u + at to calculate time when the initial and final velocities are known.
  • Use s = ut + ½at² to calculate displacement or solve for time directly.
  • At a given height during vertical motion, the velocity can have two signs: positive while rising and negative while falling.
  • The speed at the same height is equal on the upward and downward journeys when air resistance is neglected.
Key Idea: The same height can correspond to two different times because vertical motion reverses direction at the maximum height.
9

Distance Travelled During Successive Seconds in Free Fall

Question:

(II) For an object falling freely from rest, show that the distance travelled during each successive second increases in the ratio of successive odd integers (1, 3, 5, etc.). This was first shown by Galileo.

Given

  • Initial velocity: u = 0 m/s
  • Acceleration due to gravity: g = 9.80 m/s²
  • Motion is freely falling under gravity.
  • Air resistance is neglected.

Understanding the Problem

An object falling freely from rest starts with zero velocity. As it falls, its velocity increases uniformly because the acceleration due to gravity is constant.

We need to determine the distance travelled during each individual second, not the total distance travelled from the beginning of the fall.

For example:

  • First second: distance travelled between t = 0 s and t = 1 s.
  • Second second: distance travelled between t = 1 s and t = 2 s.
  • Third second: distance travelled between t = 2 s and t = 3 s.

Method 1: Using the Equations of Motion

Since the object starts from rest, its initial velocity is zero. The second equation of motion gives the total distance fallen after time t:

s = ut + ½gt²

Since u = 0:

s = ½gt²

This equation gives the total distance travelled from the starting point up to time t. We can use it to calculate the distance travelled during each successive second by subtracting the total distances at consecutive integer times.

Distance Travelled During the First Second

The distance travelled during the first second is the total distance at t = 1 s minus the distance at t = 0 s.

At t = 1 s:

s₁ = ½g(1)²

s₁ = g/2

Therefore, the distance travelled during the first second is

s₁ = g/2

Distance Travelled During the Second Second

The distance travelled during the second second is the total distance at t = 2 s minus the total distance at t = 1 s.

Total distance after 2 seconds:

s(2) = ½g(2)²

s(2) = 2g

Total distance after 1 second:

s(1) = ½g(1)²

s(1) = g/2

Distance during the second second:

d₂ = s(2) − s(1)

d₂ = 2g − g/2

d₂ = 3g/2

Distance Travelled During the Third Second

Similarly, the distance travelled during the third second is the total distance after 3 seconds minus the total distance after 2 seconds.

Total distance after 3 seconds:

s(3) = ½g(3)²

s(3) = 9g/2

Distance during the third second:

d₃ = s(3) − s(2)

d₃ = 9g/2 − 2g

d₃ = 5g/2

Comparing the Distances

We have obtained the distances travelled during the first three successive seconds:

d₁ = g/2

d₂ = 3g/2

d₃ = 5g/2

Taking the common factor g/2:

d₁ : d₂ : d₃ = 1 : 3 : 5

Therefore, the distances travelled during the first, second, and third seconds are in the ratio 1 : 3 : 5.

Method 2: Deriving the General Formula

We can prove the result for any successive second, rather than calculating each second separately.

Let s(t) represent the total distance travelled after time t. For an object falling from rest:

s(t) = ½gt²

The distance travelled during the nth second is the difference between the total distances at t = n seconds and t = (n − 1) seconds.

dₙ = s(n) − s(n − 1)

Substituting the distance equation:

dₙ = ½gn² − ½g(n − 1)²

Taking ½g as a common factor:

dₙ = ½g[n² − (n − 1)²]

Expanding the square:

(n − 1)² = n² − 2n + 1

Therefore:

dₙ = ½g[n² − (n² − 2n + 1)]

Simplifying:

dₙ = ½g(2n − 1)

dₙ = (g/2)(2n − 1)

This is the general formula for the distance travelled during the nth second by an object falling freely from rest.

Since g/2 is constant, the distance during each successive second is proportional to the factor (2n − 1).

Successive Odd Integers

  • For n = 1: 2(1) − 1 = 1
  • For n = 2: 2(2) − 1 = 3
  • For n = 3: 2(3) − 1 = 5
  • For n = 4: 2(4) − 1 = 7
  • For n = 5: 2(5) − 1 = 9

Hence, the distances travelled during successive seconds follow the ratio:

1 : 3 : 5 : 7 : 9 : ...

The ratio consists of consecutive odd positive integers.

Numerical Verification Using g = 9.80 m/s²

To verify the result numerically, we can calculate the distance travelled during each of the first five seconds.

Using the general formula:

dₙ = (9.80/2)(2n − 1)

dₙ = 4.90(2n − 1) m

Second Odd Integer Distance During That Second
1st 1 4.90 m
2nd 3 14.7 m
3rd 5 24.5 m
4th 7 34.3 m
5th 9 44.1 m

For example, during the fourth second:

d₄ = 4.90(2 × 4 − 1)

d₄ = 4.90 × 7 = 34.3 m

Notice that these are distances travelled during individual seconds, not total distances measured from the beginning of the fall.

Why Do the Distances Follow Odd Integers?

The distance travelled from rest increases with the square of time:

s ∝ t²

The total distances after 1, 2, 3, 4, and 5 seconds are proportional to:

1² : 2² : 3² : 4² : 5²

1 : 4 : 9 : 16 : 25

To find the distance travelled during each individual second, subtract consecutive square numbers:

First second: 1² − 0² = 1

Second second: 2² − 1² = 3

Third second: 3² − 2² = 5

Fourth second: 4² − 3² = 7

Fifth second: 5² − 4² = 9

In general, the difference between consecutive squares is:

n² − (n − 1)² = 2n − 1

Since 2n − 1 gives successive odd integers, the distances travelled during successive seconds follow the ratio 1 : 3 : 5 : 7 : 9 : ...

Final Results

  • Distance travelled after time t: s = ½gt².
  • Distance travelled during the nth second: dₙ = (g/2)(2n − 1).
  • The distances travelled during successive seconds are proportional to successive odd integers.
  • The resulting ratio is 1 : 3 : 5 : 7 : 9 : ....

Key Idea

For an object falling freely from rest under constant gravitational acceleration, the distance travelled during the nth second is

dₙ = (g/2)(2n − 1)

Therefore, the distances travelled during successive seconds are in the ratio

1 : 3 : 5 : 7 : 9 : ...

10

Rocket Motion with Powered Ascent, Free Ascent, and Free Fall

Question:

(II) A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 775 m. After this point, its acceleration is that of gravity, downward.

(a) What is the velocity of the rocket when it runs out of fuel?

(b) How long does it take to reach this point?

(c) What maximum altitude does the rocket reach?

(d) How much time (total) does it take to reach maximum altitude?

(e) With what velocity does it strike the Earth?

(f) How long (total) is it in the air?

Given

  • Initial velocity: u = 0 m/s
  • Acceleration during powered ascent: a₁ = +3.2 m/s²
  • Height when fuel runs out: h = 775 m
  • Acceleration after fuel runs out: a₂ = −9.8 m/s²
  • Gravitational acceleration: g = 9.8 m/s²
  • Initial position: y₀ = 0 m

Understanding the Two Stages of Motion

The rocket experiences two different stages of motion. We must treat them separately because the acceleration changes when the fuel runs out.

Stage 1: Powered Ascent

The rocket starts from rest and accelerates upward at 3.2 m/s² until it reaches 775 m.

a₁ = +3.2 m/s²

Stage 2: After Fuel Runs Out

The rocket continues upward under gravity alone, slows to zero velocity at its maximum altitude, and then falls back toward Earth.

a₂ = −9.8 m/s²

(a) Velocity When the Rocket Runs Out of Fuel

During the powered ascent, the rocket starts from rest and accelerates uniformly through a vertical displacement of 775 m. We know the initial velocity, acceleration, and displacement. Therefore, we can use the third equation of motion without calculating time first.

Method 1: Using the Third Equation of Motion

v² = u² + 2as

Substituting the values:

v² = 0² + 2(3.2)(775)

v² = 6.4 × 775

v² = 4960

Taking the positive square root because the rocket is moving upward:

v = √4960

v ≈ 70.4 m/s

The positive sign indicates upward motion. The rocket is moving upward when its fuel runs out.

Answer for (a)

The velocity of the rocket when it runs out of fuel is

+70.4 m/s (upward)

(b) Time Taken to Reach the Fuel Cutoff Point

During the first stage, the rocket starts from rest and accelerates at 3.2 m/s². We have already calculated its velocity at the end of this stage.

Method 1: Using the First Equation of Motion

v = u + at

Rearranging:

t = (v − u)/a

Substituting:

t₁ = (70.4 − 0)/3.2

Using the unrounded velocity gives:

t₁ ≈ 22.0 s

Thus, the rocket takes approximately 22.0 seconds to reach an altitude of 775 m and run out of fuel.

Method 2: Using the Second Equation of Motion

We can independently calculate the same time using the displacement equation.

s = ut + ½at²

775 = 0(t) + ½(3.2)t²

775 = 1.6t²

t² = 775/1.6

t = √484.375

t₁ ≈ 22.0 s

Answer for (b)

The time taken to reach the fuel cutoff point is

t₁ ≈ 22.0 s

(c) Maximum Altitude Reached by the Rocket

After the fuel runs out, the rocket continues moving upward because it already has an upward velocity of approximately 70.4 m/s. Gravity acts downward and reduces this velocity until it becomes zero at the highest point.

We must calculate the additional height gained after the fuel cutoff and add it to the initial 775 m altitude.

Step 1: Calculate the Additional Height

At the maximum altitude, the final velocity is zero. Taking the fuel cutoff point as the starting position for this stage, we use the third equation of motion.

v² = u² + 2as

For this stage:

u = +70.4 m/s, v = 0, a = −9.8 m/s²

0 = u² − 2gs

Rearranging:

s = u²/(2g)

Using the unrounded cutoff velocity:

s = 4960/[2(9.8)]

s = 4960/19.6

s ≈ 253.1 m

Step 2: Calculate the Maximum Altitude

The rocket has already reached 775 m when its fuel runs out. It rises an additional 253.1 m before stopping momentarily.

H = 775 + 253.1

H ≈ 1028.1 m

The maximum altitude is measured from the Earths surface, which is taken as the zero-height reference.

Answer for (c)

The maximum altitude reached by the rocket is

H ≈ 1.028 × 10³ m

(d) Total Time to Reach Maximum Altitude

The total ascent time consists of two parts:

  • Time during powered ascent.
  • Additional time after the fuel runs out, while the rocket continues upward under gravity.

Step 1: Calculate the Additional Ascent Time

During the second stage, the initial velocity is approximately 70.4 m/s and the final velocity at the highest point is zero. The acceleration is −9.8 m/s².

v = u + at

0 = u − gt

gt = u

t₂ = u/g

Substituting the unrounded velocity:

t₂ = √4960/9.8

t₂ ≈ 7.19 s

Step 2: Calculate the Total Ascent Time

T = t₁ + t₂

T = 22.0 + 7.19

T ≈ 29.2 s

The rocket takes approximately 29.2 seconds from launch to reach its maximum altitude.

Answer for (d)

T ≈ 29.2 s

(e) Velocity with Which the Rocket Strikes the Earth

After reaching its maximum altitude, the rocket falls freely under gravity. We can calculate its impact velocity using the third equation of motion, starting from the highest point.

At the maximum altitude:

  • Initial velocity for the falling stage: u = 0 m/s
  • Displacement to the ground: s = −1028.1 m
  • Acceleration: a = −9.8 m/s²

Method 1: Using the Third Equation of Motion

v² = u² + 2as

Substituting:

v² = 0² + 2(−9.8)(−1028.1)

Using the unrounded maximum altitude:

v² ≈ 20150

Taking the square root:

v ≈ ±142.0 m/s

The positive root would represent upward motion, whereas the negative root represents downward motion. Since the rocket is striking the Earth while falling, we choose the negative root.

Alternative Check: Starting at Fuel Cutoff

We can also calculate the impact velocity directly from the fuel cutoff point, where the rocket is moving upward at approximately 70.4 m/s from an altitude of 775 m.

v² = u² + 2as

v² = 4960 + 2(−9.8)(−775)

v² = 4960 + 15190

v² = 20150

v ≈ −142.0 m/s

The negative sign indicates that the rocket is moving downward when it reaches the ground.

Answer for (e)

Taking upward as positive, the velocity when the rocket strikes the Earth is

v ≈ −142.0 m/s

Speed of impact ≈ 142.0 m/s, directed downward.

(f) Total Time the Rocket Is in the Air

The total time in the air consists of three intervals:

  • Powered ascent.
  • Additional ascent after the fuel runs out.
  • Descent from the maximum altitude to the ground.

Step 1: Calculate the Time Taken to Fall from Maximum Altitude

From the maximum altitude, the rocket starts falling from rest. We use the second equation of motion to calculate the falling time.

s = ut + ½at²

Taking downward as the direction of displacement for this falling stage:

H = ½gt₃²

Rearranging:

t₃ = √(2H/g)

Substituting H ≈ 1028.1 m:

t₃ = √[2(1028.1)/9.8]

t₃ = √209.8

t₃ ≈ 14.5 s

Step 2: Calculate the Total Time in the Air

The total time is the sum of the powered ascent time, the additional ascent time, and the falling time.

T_air = t₁ + t₂ + t₃

T_air ≈ 22.0 + 7.19 + 14.5

T_air ≈ 43.7 s

Therefore, the rocket remains in the air for approximately 43.7 seconds from launch until it reaches the Earth.

Answer for (f)

T_air ≈ 43.7 s

Summary of All Six Answers

Part Quantity Answer
(a) Velocity at fuel cutoff +70.4 m/s
(b) Time to fuel cutoff 22.0 s
(c) Maximum altitude 1028.1 m
(d) Total time to maximum altitude 29.2 s
(e) Impact velocity −142.0 m/s
(f) Total time in the air 43.7 s

Key Equations

  • Third equation of motion: v² = u² + 2as
  • First equation of motion: v = u + at
  • Second equation of motion: s = ut + ½at²
  • Additional upward height after fuel cutoff: h = u²/(2g)
  • Time to maximum altitude after fuel cutoff: t = u/g
  • Time to fall from maximum altitude: t = √(2H/g)

Key Idea

When solving a multistage motion problem, calculate the velocity and time at the transition between stages. Use the final conditions of one stage as the initial conditions of the next stage.

Powered ascent → Free ascent → Free fall

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