Kinematics Question 176
Question: A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact it’s [BHU 1997; CPMT 1997]
Options:
A) 2100 $ m/{{\sec }^{2}} $ downwards
B) 2100 $ m/{{\sec }^{2}} $ upwards
C) 1400 $ m/{{\sec }^{2}} $
D) 700 $ m/{{\sec }^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Velocity at the time of striking the floor,
$ u=\sqrt{2gh _{1}}=\sqrt{2\times 9.8\times 10}=14m/s $ Velocity with which it rebounds.
$ v=\sqrt{2gh _{2}}=\sqrt{2\times 9.8\times 2.5}=7\ m/s $
$ \therefore $ Change in velocity $ \Delta v=7-(-14)=21m/s $
$ \therefore $ Acceleration $ =\frac{\Delta v}{\Delta t}=\frac{21}{0.01}=2100\ m/s^{2} $ (upwards)
