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.