Question

5. A body falling from a vertical height of 10 m pierces
through a distance of 1 m in sand. It faces an average
retardation in sand equal to (g = acceleration due to
gravity)

-1) g
2) 100g
(3) 9g
(4) 1000g​

Answer 1

Answer:

Given,

u=0

S = 10 m

S=1/2gt^2

10 = 1/2×10×t^2

5t^2= 10

t=√2

v = u+at

= 10√2

For the second part of the question,

u =v

a=?

S=1

V=0

V^2- v^2 / 2S. = a

-200/2 = a

a =-100

g=10 m/s^2

a in terms of g is-10g

Hope it helps and please mark as brainliest:)