A ball is dropped from a height of 20 m and rebounds with a velocity which is 3/4th of the velocity with which it hits the ground. What is the time interval between the first and second bounces (g = 10 m/s 2 )
Text Solution
Verified by ExpertsThe correct answer is:
A
V 1 = 
= 
= 20m/sec
V 2 = 
× 20 = 15m/sec
t 2 = 
= 
= 1.5 sec
total time = 2 × 1.5 = 3 sec
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A particle moving in a straight line covers half the distance with speed of 3 m/s . The other half …
A particle starts from rest. Its acceleration ( a ) versus time ( t ) is as shown in the figure. Th…
A car accelerates from rest at a constant rate for some time, after which it decelerates at a cons…
A stone dropped from a building of height and it reaches after seconds on earth. From the same bu…
A ball is projected upwards from a height above the surface of the earth with velocity . The time…
A particle is dropped vertically from rest from a height. The time taken by it to fall through succ…