Answer :

A ball is dropped from a height of 5 m (s) above the sand level.



The same ball penetrates the sand up to 10 cm (
) before coming to rest.
Initial velocity of the ball, u = 0
and, a = g = 9.8 m/s2 (Acceleration due to gravity)


Using the equation of motion, we get:





t=1.01 s


Thus, the time taken by the ball to cover the distance of 5 m is 1.01 seconds.



Velocity of the ball after 1.01 s:
v = u + at
v = 9.8 × 1.01 = 9.89 m/s
Hence, for the motion of the ball in the sand, the initial velocity u2 should be 9.89 m/s and the final velocity v2 should be 0.


= 10 cm = 0.1 m



Again using the equation of motion, we get:



Hence, the sand offers the retardation of 490 m/s2.


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