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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.



Answer :

avtrisal
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.
Edufirst

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

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