Q. 304.0( 10 Votes )

# A ball is dropped

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/s^{2} (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 *u*_{2} should be 9.89 m/s and the final velocity *v*_{2} 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/s^{2}.

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