A golf ball is dropped from rest from a height of 9.5m. It hits the pavement then bounces back up rising just 5.7 m before falling back down again. A boy then catches the ball on the way down when it is 1.20 m above the pavement. Ignoring air resistance, calculate the total amount of time the ball is in the air, from drop to catch.

The time the ball takes to fall 9.5 meters is the square root of (19/g), where g is gravitational acceleration.
The time it takes to rise to 5.7 meters is the square root of (11.4/g), for the same value of g.
The time it takes to fall from 5.7 meters to 1.2 is the square root of (9/g).
So the answer is [sqrt(19)+sqrt(11.4)+sqrt(9)]/sqrt(g). If g=10, the answer is 3.39 seconds; if g=9.8, the answer is 3.43 seconds.

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Edufirst Edufirst · Answer 2 · 2018-08-23T00:33:02-04:00

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

1) Ball dropping from 9.5m: free fall

d = Vo + gt² / 2

Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²

⇒ t = √ (1.94 s²) = 1.39s

2) Ball rising 5.7m (vertical rise)

i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

4) Total time

t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s