A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. Calculate the average retardation in sand faced by the body. ( g= acceleration due to gravity)

 a)g

 b)9g

 c)100g

 d)1000g

When the body falls from the height :

u = 0,

s = 10 m ,

Using the equation v²=u²+2as

v² = 0 + 2g×10

v = √20g

The velocity when it touches the sand is √20g

So when the body pierces through sand : u = √20g, v=0, s = 1 m

v²=u²+2as

0=(√20g)²+2a

-a = 20g / 2

-a = 10 g

The average retardation in sand faced by the body is 10 g