Question

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

Answer 1

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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Hence the correct alternative is

(1) 0.09 sec