The body falls from rest (u=0) through a distance h before striking the ground, the speed v of the body is given by kinematical equation:v2=u2+2as;
Putting a=g and s=h, we obtain v=√2gh
Thus, the magnitude of the total change in momentum of the body just before it strikes the ground=ΔP=|mv−0|=mv
Where v=√2gh⇒ΔP=m√2gh
or ΔP=(1)√(2×10×20)kgms−1
=20kgms−1......(i)
We define average force experienced by the body,
Fav−−→=ΔPΔt;Δt= time of motion of the body = t(say).
We know ΔP=20kgms−1 from (i). Now, let us find t using the facts given in problem. From kinematics, we know
s=ut+12at2 (here u=0,s=h, and a=g)
⇒h=12gt2 or t=√2hg
∴ Fav=ΔPΔt=ΔPt
Putting the general values of ΔP and t, we get
Fav=m√2gh√2h/g=mg
⇒Fav=mg
Where mg is the weight of the body and g is directed vertically downward. Therefore, the body experiences a constant vertically downward force of magnitude mg.