Question

a stone weighing 3kg falls from the top of a tower 100m high and buries itself 2m deep in the sand .What is the time of penetration ?​

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 answer is =0.09s

Answer 2

Given :

m = 3 kg

u = 0

h = 100 m

g = 9.8 m/s²

Using 3rd equation of motion :

v²-u² = 2gh

v²-0² = 2 (9.8) 100

v = 2√490

v = 14√10 m/s

___________________________

Now, during penetration :

u = 14√10 m/s

v = 0

S = 2 m

a = ?

t = ?

Using 3rd equation of motion;

v²-u² = 2aS

0²-(14√10)² = (2) a (2)

4a = -1960

a = -490 m/s²

------------------------------------------

Now, using 1st equation is motion ;

v = u+at

0 = 14√10 + (-490)t

t = 14√10/490

t = 0.0903 sec

===========================

Hence the correct alternative is

(1) 0.09 sec