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)



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