Question

A body falling from a vertical height of 10m pierces through a distance of 1m in sand.it faces an average retardation equal to?

Answer 1

Given :
h=10m
For free fall :
Initial velocity=u=0 m/s
Let v be the final velocity.
From third equation of motion :
V²-u²=2ah
a=g=10m/s²
v=√(2x10x10) =√200m/s

Now for motion in sand ,
From third equation of motion :
V1²-u²=2as
Final velocity =V1= in sand = 0 m/s
u = v
s=1m
v1²-u²=2as
0-²(√200)²=2a(1)
-200=2a
a=-100/2
=-100m/s²

Negative sign shows the retardation of the ball in sand is 100m/s2=10g
So, Retardation=10g

Answer 2

Answer:

The correct answer is 9g . Hope this helps ☺