A body of mass m=1kg falls from a height h=20m from the ground level .
(a) What is the magnitude of total change in momentum of the body before it strikes the ground?
(b) What is the corresponding average force experienced by it ?`(g=10ms^(-1))`

Explanation

The body falls rest `(u=0)` through a distance h before striking the ground, the speed v of the body is given by kinematical equation: `V^(2)=u^(2)+2as`:
Putting `1-g` and `s=h`, we obtain `v=sqrt(2gh)`
Thus, the magnitude of the total change in momentum of the body just before it strikes the ground `=DeltaP=|mv-0|=mv`
where `v=sqrt(2gh)implies Delta P=msqrt(2gh)`
or, `DeltaP =(1)sqrt((2xx10 xx 20))kg ms^(-1)`
`=20 kgms^(-1)` ...(i)
we define average force experienced by the body
`vec(F_(av))=(Delta vec(P))/(Delta t): Delta t=`time of motion of the body =t (say).
we know `DeltaP=20 kg ms^(-1)` from (i) Now let us find t using the facts given in problem. from kinematics, we know
`s=ut+1/2 at^(2)` (hare u=0, s=h and a=g)
`implies h=1/2 g t^(2)` or `t=sqrt((2h)/(g))`
`:. F_(av)=(Deltap)/(Delta t)=(Delta p)/(t)`
Putting the general values of `Delta p and t`, we get
`F_(av)=(msqrt(2gh))/(sqrt(2h//g))=mg`
`implies vec(F)_(av)=m vec(g)`
where mg is the weight of the body and `vec(g)` is directed vertically downward. Therefore the body expriences a constant vertically downward force of magnitude mg.