Question

a stone of mass 5kg falls from the top of a cliff 50m high and buries 1m in deep in sand.find average resistant offered by sand

Answer 1

m = 5 kg
h = 50 m

velocity = v just before hitting the ground 
 v² = u² + 2 g h = 0 + 2 * 9.8 * 50 = 980

When the stone is penetrating the sand:
v² = u² + 2 a s
0 = 980 + 2 * a * 1
a = - 490 m/s²

Average resistance Force offered by sand = m a = - 5 * 490
   =- 2, 450 N

Answer 2

Answer:

Explanation:M = 5 kg

h = 50 m

velocity = v just before hitting the ground 

Using third equation of motion

 v² = u² + 2 g h = 0 + 2 * 9.8 * 50

=980

When the stone is penetrating the sand:

v² = u² + 2 a s

0 = 980 + 2 * a * 1

a = - 490 m/s²

Average resistance Force offered by sand = m a = - 5 * 490

  =- 2, 450 N