Given, mass of body
=M Height above the sand floor
=h Distance penetrated into sand
=x Let
v be the velocity with which body strikes the surface.
Then, by equation of motion, we get
v2=u2+2gh⇒v2=2gh ...(i)
[∵ Initial velocity at highest point,
u=0] With this velocity
v, when body passes through the sand floor it comes to rest after travelling a distance
x.
Let
F be the resisting force acting on body.
Then, net downward force will be
Fnet=Mg−F By work-energy theorem, work done by all forces is equal to change in
KE, which gives
W=ΔK Fnet×x=Kf−Ki (Mg−F)x=0−Mv2 [∵Final‌KE,Kf=0] (Mg−F)x=−Mv2 ...(ii)
Substituting the value from Eq. (i) in Eq. (ii), we get
⇒Mg‌x−Fx=−M(2gh) ⇒Mg‌x−Fx=−Mg‌h ⇒Fx=Mg‌x+Mg‌h ⇒F=Mg()