Question

A STONE WEIGHING 3KG FALLS FROM THE TOP OF THE TOWER 100M HIGH AND BURIES ITSELF 2M DEEP IN THE SAND . WHAT IS THE TIME OF PENETRATION .

Answer 1

the potential energy at the 100m = mgh = 3 × 100 × 10 = 3000J

the kinetic energy before penetration is 1/2 mv^2 = mgh at 100m above

1/2 mv^2 = 3000

v^2 = 2000
v = √2000 = 20√5m/s

during penetration

s = 2 m
v = 0 m/s
u = 20√5 m/s

v^2 = u^2 + 2as
0 = 2000 + 2a×2
2000 /4 = -a
500= -a
a = - 500 m/s^2

v = u + at
0 = 20√5 -500× t
20√5 = 500t
t = √5/25 = 0.089 seconds

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

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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

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Now, using 1st equation is motion ;

v = u+at

0 = 14√10 + (-490)t

t = 14√10/490

t = 0.0903 sec

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(1) 0.09 sec