Question

A stone weighing 5kg. falls from the top of a tower 100m high and buries itself 1 m deep in the sand. What is the average resistance offered by sand?

Answer 1

Mass of the stone, m = 5kg

Height of the tower h = 

s = 100m

Initial velocity u =0

Final velocity v = ?

From the relation, v² = u² + 2gs

v² = 0 + 2 × 9.8 × 100

v² = 1960

V =√1960

v = 44.27 ms^-1

Then the stone penetrates through the sand with a initial velocity, 

u = 44.27 ms^-1

Distance travelled, S = 1 m

Final velocity, v = 0

acceleration, a =?

From the equation, V² = u²+2as

0² = (44.27)² + 2 × a × 1

– 1960 = 2a

a = -980ms^-2

∴ The average resistance offered by the sand is 

F = ma = 5 × 980 

F = 4900 N.