(I)A stone is dropped from rest from the top of a cliff. It is observed to hit the ground after 3 seconds. How high is the cliff? Ignore air resistance and take g = 10 m/s².
Explanation
The stone is dropped from rest, so u = 0
Using the equation of motion: h = ut + ½gt²
h = (0 × 3) + (½ × 10 × 3²)
h = 0 + 45
h = 45 m
2
(I)Estimate (a) how long King Kong would take to fall vertically from the top of the Empire State Building, 500 m high, and (b) his velocity just before landing. Assume he starts from rest and ignore air resistance. (Take g = 10 m/s².)
Explanation
(a) Using h = ut + ½gt², with u = 0:
h = ½gt²
t = √(2h/g)
t = √((2 × 500) / 10)
t = 10 s
(b) Using v = u + gt:
v = 0 + (10 × 10)
v ≈ 100 m/s downward
3
(II)A ball player catches a ball 1 seconds after throwing it vertically upward. Assuming the ball returns to the same height from which it was thrown, determine (a) the initial speed of the ball and (b) the maximum height it reaches. Ignore air resistance. (Take g = 10 m/s².)
Explanation
The ball returns to the same height, so total flight time T = 2u/g
Therefore, initial speed u = gT/2
u = (10 × 1) / 2 = 5 m/s
At maximum height, final velocity v = 0
Using v² = u² − 2gh, maximum height H = u²/2g
H = (5)² / (2 × 10) = 1.25 m
4
(II)A baseball is hit almost vertically upward with an initial speed of 50 m/s. Estimate (a) the maximum height it reaches and (b) the total time it remains in the air before returning to the launch height. (c) What factors make these estimates rather than exact predictions? Ignore air resistance for the calculations and take g = 10 m/s².
Explanation
(a) At maximum height, final velocity v = 0
Using v² = u² − 2gH
H = u² / 2g
H = (50)² / (2 × 10) = 125 m
(b) Time to reach maximum height: t = u/g
Total time of flight: T = 2u/g
T = (2 × 50) / 10 = 10 s
(c) These are estimates because air resistance, wind, the exact launch angle, variations in gravity, and the ball’s spin can affect its motion.
5
(II)A kangaroo jumps vertically upward and reaches a maximum height of 20 m. How long does it remain in the air before returning to the ground? Ignore air resistance and take g = 10 m/s².
Explanation
At maximum height, final velocity v = 0
Using v² = u² − 2gh
u² = 2gh
u = √(2 × 10 × 20) = 20 m/s
Time to reach maximum height: t = u/g
t = 20/10 = 2 s
Total time of flight: T = 2t
T = 2 × 2 = 4 s
6
(II)The best rebounders in basketball can achieve a vertical leap of about 120 cm. Estimate (a) their initial launch speed off the ground and (b) the total time they remain in the air. Assume they land at the same height from which they jumped and ignore air resistance. (Take g = 9.8 m/s².)
Explanation
(a) At maximum height, final velocity v = 0
Using v² = u² − 2gh
u² = 2gh
u = √(2 × 9.8 × 1.2)
u = 4.85 m/s
(b) Time to reach maximum height: t = u/g
Total time in air: T = 2u/g
T = (2 × 4.85) / 9.8 = 0.99 s
7
(II)A stone is thrown vertically upward with an initial speed of 50 m/s. (a) How fast is it moving when it reaches a height of 125 m? (b) How much time is required to reach this height? (c) Why are there two answers to part (b)? Take g = 10 m/s² and ignore air resistance.
Explanation
(a) Using the equation of motion v² = u² − 2gh:
v² = (50)² − 2 × 10 × 125
v = √(50² − 2 × 10 × 125)
Speed at 125 m = 0 m/s
(b) Using the equation of motion h = ut − ½gt²:
½gt² − ut + h = 0
Solving the quadratic equation gives two possible times:
t₁ = (u − √(u² − 2gh))/g = 5 s
t₂ = (u + √(u² − 2gh))/g = 5 s
(c) The stone passes the same height twice: once while moving upward and again while moving downward.
8
(II)An object falls freely from rest under gravity. Show that the distances travelled during successive one-second intervals are in the ratio of successive odd integers (1:3:5:...). Calculate the distances for the first 2 seconds and identify their ratio. Take g = 10 m/s².
Explanation
Since the object starts from rest, initial velocity u = 0.
Using the equation of motion: s = ut + ½gt²
Since u = 0, the distance fallen after t seconds is s = ½gt².
Distance travelled during the nth second:
sₙ = s(n) − s(n − 1)
sₙ = ½gn² − ½g(n − 1)²
sₙ = ½g[n² − (n − 1)²]
Expanding: sₙ = ½g(2n − 1)
Therefore, the distance during each successive second is proportional to (2n − 1).
For n = 1, 2, 3, 4, ... the factors are 1, 3, 5, 7, ...
For g = 10 m/s², the distances during the first 2 seconds are: 5 m, 15 m.
Their ratio is 1:3.
9
(II)A rocket rises vertically from rest with a constant upward acceleration of 10 m/s² until it runs out of fuel at an altitude of 500 m. After this point, its acceleration is 10 m/s² downward due to gravity. Neglect air resistance. Calculate: (a) its velocity when the fuel runs out, (b) the time taken to reach that point, (c) its maximum altitude, (d) the total time from launch to maximum altitude, (e) its velocity when it strikes Earth, and (f) the total time it remains in the air.
Explanation
Take upward as positive and use g = 10 m/s² downward.
(a) Powered ascent: u = 0, a = 10 m/s², s = 500 m.
Using v² = u² + 2as:
v² = 0 + 2 × 10 × 500
v = 100 m/s upward.
(b) Using v = u + at:
t₁ = (100 − 0) / 10
t₁ = 10 s.
(c) After fuel exhaustion, the rocket continues upward while gravity slows it.
At maximum altitude, v = 0.
Using v² = u² − 2gh_coast:
h_coast = v_fuel² / (2g) = 500 m.
Maximum altitude = 500 + 500 = 1000 m.
(d) Time from fuel exhaustion to maximum altitude:
Using v = u − gt, with final velocity zero:
t₂ = v_fuel / g = 10 s.
Total time to maximum altitude = 10 + 10 = 20 s.
(e) The rocket falls from rest at maximum altitude to Earth.
Using v² = u² + 2gh_max:
v = √(2 × 10 × 1000) = 141.42 m/s downward.
(f) During the fall from maximum altitude:
Using h_max = ½gt²:
t₃ = √(2h_max / g) = 14.14 s.
Total time in the air = t₁ + t₂ + t₃ = 34.14 s.
10
(II)A helicopter is ascending vertically with a speed of 50 m/s. At a height of 500 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? Take g = 10 m/s² and neglect air resistance.
Explanation
When the package is released, it retains the upward velocity of the helicopter.
Initial velocity: u = 50 m/s upward.
Initial height: h = 500 m.
Acceleration: a = −10 m/s², taking upward as positive.
Using the equation of motion: y = h + ut − ½gt².
At the ground, y = 0.
Therefore, 0 = h + ut − ½gt².
Rearranging: ½gt² − ut − h = 0.
Using the quadratic formula: t = (u + √(u² + 2gh)) / g.
t = (50 + √(50² + 2 × 10 × 500)) / 10.
t = 16.18 s. The positive root is chosen because time must be positive.