Question

A golf ball is dropped from rest from a height of 8.30 m. It hits the pavement, then bounces back up, rising just 6.00 m before falling back down again. A boy then catches the ball when it is 1.20 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

Answer 1

time for the fall from 9.50m to the pavement is t1,

h = (1/2)gt1²

=>t1 = square root(2h/g)

= square root(2*8.30m/9.8m/s²)= 1.30s

since the rebound max height is 6.0 m, the rebound speed vis:

v² /2g = h1 = 6.0 m

=>v = square root (2*6.0m*9.8m/s²) = 10.84m/s

the time to reach max heigh from rebound t2 = v/g = 10.84 m/s /9.8m/s² = 1.11s

the time to fall again and catch t3 is:

h2 = (6.0m-1.20m) = (1/2)g(t3)²

=>t3 = square root [(6.0m-1.20m)*2/9.8m/s²] =0.99s

the total time is:

t = t1+t2+t3 = 1.30s + 1.11s+0.99s = 3.4 s

answer:3.4 s

I hope help you !!