Answer :
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.
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.
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
