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Freely Falling Objects - Examples

1

Falling from a Tower: Distance Fallen at Different Times

Falling from a tower

Suppose that a ball is dropped (v0 = 0) from a tower. How far will it have fallen after a time t1 = 1.00 s, t2 = 2.00 s, and t3 = 3.00 s? Ignore air resistance.

Given data

  • Initial velocity: u = 0 m/s
  • Acceleration due to gravity: g = 9.80 m/s2
  • First time: t1 = 1.00 s
  • Second time: t2 = 2.00 s
  • Third time: t3 = 3.00 s

Step 1: Identify the type of motion

The ball is released from rest and falls freely under gravity. Air resistance is neglected, so the ball has constant downward acceleration.

Taking downward as the positive direction, we use the equation of motion:

s = ut + ½at2

Here, u = 0 and a = g.

s = ½gt2

Step 2: Distance fallen after 1.00 s

Substitute t = 1.00 s:

s1 = ½ × 9.80 × (1.00)2

s1 = 4.90 m

Distance fallen = 4.90 m

Step 3: Distance fallen after 2.00 s

Substitute t = 2.00 s:

s2 = ½ × 9.80 × (2.00)2

s2 = 4.90 × 4

s2 = 19.6 m

Distance fallen = 19.6 m

Step 4: Distance fallen after 3.00 s

Substitute t = 3.00 s:

s3 = ½ × 9.80 × (3.00)2

s3 = 4.90 × 9

s3 = 44.1 m

Distance fallen = 44.1 m

Final answers

Time Distance fallen
1.00 s 4.90 m
2.00 s 19.6 m
3.00 s 44.1 m

Key idea

For an object dropped from rest under constant gravitational acceleration, the distance fallen is proportional to the square of elapsed time:

s ∝ t2

Therefore, doubling the time makes the distance four times greater, while tripling the time makes the distance nine times greater.

2

Thrown Down from a Tower

Thrown down from a tower

Suppose the ball is thrown downward with an initial velocity of 3.00 m/s, instead of being dropped. (a) What then would be its position after 1.00 s and 2.00 s? (b) What would be its speed after 1.00 s and 2.00 s? Compare with the speeds of a dropped ball.

Given data

  • Initial downward velocity: u = 3.00 m/s
  • Acceleration due to gravity: g = 9.80 m/s2
  • First time: t1 = 1.00 s
  • Second time: t2 = 2.00 s
  • Air resistance is neglected.

Sign convention

Take the downward direction as positive and the tower’s top as the origin. The ball moves downward with an initial velocity of 3.00 m/s, and gravity also acts downward.

u = +3.00 m/s

a = +9.80 m/s2

Step 1: Select the equations of motion

Since the acceleration is constant, we can use the equations of motion.

y = ut + ½at2

v = u + at

Part (a): Position after 1.00 s and 2.00 s

Step 2: Position after 1.00 s

Using y = ut + ½at2:

y1 = 3.00(1.00) + ½(9.80)(1.00)2

y1 = 3.00 + 4.90

y1 = 7.90 m

After 1.00 s, the ball is 7.90 m below its starting point.

Step 3: Position after 2.00 s

y2 = 3.00(2.00) + ½(9.80)(2.00)2

y2 = 6.00 + 19.60

y2 = 25.6 m

After 2.00 s, the ball is 25.6 m below its starting point.

Part (b): Speed after 1.00 s and 2.00 s

Use the velocity equation:

v = u + at

Step 4: Speed after 1.00 s

v1 = 3.00 + 9.80(1.00)

v1 = 12.80 m/s

Speed = 12.8 m/s

Step 5: Speed after 2.00 s

v2 = 3.00 + 9.80(2.00)

v2 = 22.60 m/s

Speed = 22.6 m/s

Comparison with a dropped ball

For a ball dropped from rest, the initial velocity is zero. Its position and speed are given by

y = ½gt2

v = gt

Time Dropped ball: position Thrown ball: position Dropped ball: speed Thrown ball: speed
1.00 s 4.90 m 7.90 m 9.80 m/s 12.8 m/s
2.00 s 19.6 m 25.6 m 19.6 m/s 22.6 m/s

Final answers

(a) Position measured downward from the starting point

After 1.00 s: 7.90 m

After 2.00 s: 25.6 m

(b) Speed

After 1.00 s: 12.8 m/s

After 2.00 s: 22.6 m/s

Key idea

A ball thrown downward travels farther and has a greater speed than a ball dropped from rest at the same time because it already has an initial downward velocity. Both balls have the same acceleration due to gravity when air resistance is neglected.

3

Ball Thrown Upward

A person throws a ball upward into the air with an initial velocity of 15.0 m/s. Calculate how high it goes. Ignore air resistance.

Given

  • Initial velocity, u = +15.0 m/s
  • Final velocity at maximum height, v = 0 m/s
  • Acceleration due to gravity, a = −9.80 m/s²
  • Maximum height, h = ?

Step 1: Choose the equation of motion.

v² = u² + 2as

Step 2: Substitute the known values.

0² = (15.0)² + 2(−9.80)h

0 = 225 − 19.6h

19.6h = 225

h = 225 / 19.6

Step 3: Calculate the maximum height.

h ≈ 11.5 m

Final Answer: The ball reaches a maximum height of approximately 11.5 m above its release point.

Key idea: At the highest point, the ball’s instantaneous velocity is zero, but its acceleration remains 9.80 m/s² downward.

4

Ball Thrown Upward – Time of Flight

Problem: Ball thrown upward, II. In Fig. 2–23, Example 2–12, how long is the ball in the air before it comes back to the hand?

Given

  • Initial velocity, u = 15.0 m/s
  • Final velocity, v = −15.0 m/s
  • Acceleration due to gravity, a = −9.80 m/s²
  • Total time of flight, t = ?

Step 1: Use the first equation of motion.

v = u + at

Step 2: Substitute the values.

−15.0 = 15.0 − 9.80t

−30.0 = −9.80t

t = 30.0 / 9.80

Step 3: Calculate the time.

t ≈ 3.06 s

Final Answer: The ball remains in the air for approximately 3.06 s.

Key idea: When the ball returns to the same height from which it was thrown, its velocity has the same magnitude as its initial velocity but points downward. The total time of flight is twice the time taken to reach maximum height.

5

Ball Thrown Upward – Time to Reach Maximum Height and Return Velocity

Problem: Ball thrown upward, III. Let us consider again the ball thrown upward of Examples 2–12 and 2–13, and make more calculations. Calculate (a) how much time it takes for the ball to reach the maximum height (point B in Fig. 2–23), and (b) the velocity of the ball when it returns to the thrower’s hand (point C).

In Examples 2–12 and 2–13, the ball is thrown upward with an initial velocity of 15.0 m/s. Ignore air resistance and take the acceleration due to gravity as 9.80 m/s² downward.

Given

  • Initial velocity, u = +15.0 m/s
  • Acceleration, a = −9.80 m/s²
  • Velocity at maximum height, v = 0 m/s

Part (a): Time to reach maximum height

At the maximum height, the ball momentarily stops moving upward, so its vertical velocity is zero.

v = u + at

0 = 15.0 − 9.80t

9.80t = 15.0

t = 15.0 / 9.80

t ≈ 1.53 s

Therefore, the ball takes approximately 1.53 s to reach its maximum height.

Part (b): Velocity when the ball returns to the thrower’s hand

The ball returns to its original height. Neglecting air resistance, its speed at this height is equal to its initial speed, but its direction is downward.

Taking upward as the positive direction, the velocity when the ball returns is negative.

v = −u

v = −15.0 m/s

v = −15.0 m/s (downward)

Final Answers: (a) Time to reach maximum height = 1.53 s. (b) Return velocity = 15.0 m/s downward, or −15.0 m/s when upward is positive.

Key idea: At the highest point, velocity is zero, but acceleration is still 9.80 m/s² downward. When the ball returns to its original height, it has the same speed as it had initially, but its velocity is in the opposite direction.

6

Ball Thrown Upward at the Edge of a Cliff

Ball Thrown Upward at the Edge of a Cliff

Example 2–16: Ball thrown upward at edge of cliff. Suppose that the person of Examples 2–12, 2–13, and 2–15 throws the ball upward at 15.0 m/s while standing on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below, as shown in Fig. 2–25a. (a) How long does it take the ball to reach the base of the cliff? (b) What is the total distance traveled by the ball? Ignore air resistance (likely to be significant, so our result is an approximation).

Given

  • Initial velocity: u = +15.0 m/s
  • Acceleration due to gravity: a = −9.80 m/s²
  • Final position relative to the cliff edge: y = −50.0 m
  • Initial position: y₀ = 0 m
  • Time to reach the base: t = ?

Part (a): Time taken to reach the base of the cliff

Method 1: Using the displacement equation

For constant acceleration, the vertical position is given by:

y = y₀ + ut + ½at²

Taking the cliff edge as the origin and upward as positive:

−50.0 = 0 + 15.0t − 4.90t²

4.90t² − 15.0t − 50.0 = 0

Using the quadratic formula, t = [−b ± √(b² − 4ac)] / 2a, with a = 4.90, b = −15.0, and c = −50.0:

t = [15.0 ± √(225 + 980)] / 9.80

t = [15.0 ± √1205] / 9.80

t = [15.0 ± 34.7] / 9.80

t ≈ 5.07 s or −2.01 s

The negative time is not physically relevant because we measure time from the instant the ball is thrown. Therefore, we choose the positive value.

t ≈ 5.07 s

Method 2: Using the velocity equation first

We can also find the maximum height and then calculate the time for the downward journey.

At the highest point, the velocity is zero. Using v = u + at:

0 = 15.0 − 9.80t₁

t₁ = 15.0 / 9.80 ≈ 1.53 s

The additional height reached above the cliff edge is found using v² = u² + 2as:

0 = (15.0)² − 2(9.80)h

h = 225 / 19.6

h ≈ 11.5 m

From the highest point, the ball falls a total vertical distance of:

s = 11.5 + 50.0

s ≈ 61.5 m

Starting from rest at the highest point, use s = ½gt₂²:

61.5 = 4.90t₂²

t₂ = √(61.5 / 4.90)

t₂ ≈ 3.54 s

Total time:

t = t₁ + t₂

t ≈ 1.53 + 3.54

t ≈ 5.07 s

Part (b): Total distance traveled by the ball

The ball first travels upward from the cliff edge to its maximum height, then travels downward from that highest point to the base of the cliff.

Upward distance: s₁ ≈ 11.5 m

Downward distance: s₂ ≈ 11.5 + 50.0 = 61.5 m

Total distance is the sum of the upward and downward distances:

Total distance = s₁ + s₂

= 11.5 + 61.5

Total distance ≈ 73.0 m

Final Answers: (a) Time to reach the base ≈ 5.07 s. (b) Total distance traveled ≈ 73.0 m.

Key idea: Displacement includes direction, so the final displacement is −50.0 m. Distance is the total path traveled, so it includes both the upward journey and the entire downward journey.

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