Question

a ball is dropped from a height of 20m and rebounds with a velocity which ks 3/4th of the velocity with which it hits the ground. what is the time interval between the first and second rebounds

Answer 1

let the speed of the wall from which it hits the ground = x m/s

for first case =
v²=u²+2as
here initial velocity U IS =0
SO , v² = 0+2(10)20
v = √400 = 20 m/s

velocity after ball rebound = 3/4×20

= 15 m/s
now, time interval b/w 1st and 2nd rebound is equal to the time taken by ball to reach highest point after 1st rebound + the taken to reach the ground again for 2nd rebound ..

so, during upward motion after 1st rebound-
v=u+at
here a will be -ve , (a=10m/s²)

so, 0 = 15+ -10t
t = 3/2 = 1.5 sec.
total time = 2t
= 2×1.5 = 3 sec
so, the time interval between first and second Rebounds will be 3 SEC.

⚡HERE IS YOUR ANS.⚡
✨HOPE IT HELPS U ✨

Answer 2

the correct answer is 3seconds