Question

A stone weighing 2 kg falls from the top of a tower 40 meter high and burries itself 1m deep in sand.The time of penetration is? (take g =9.8ms-1)

Answer 1

Answer:

ANSWER

The velocity of the ball before reaching the ground is given by

v

2

=u

2

+2as

v

2

=2×9.8×100

v=14

10

​

Distance penetrated before coming to rest =2m

If the retardation experienced while penetrating is a

v

2

=u

2

−2as

v=0

u

2

=2as

a=490m/s

2

retardation

v=u−at

t=

490

14

10

​

​

=0.09s