A body starting from rest slides on an inclined plane of lengths as shown in Calculate the time of descent and speed at the bottom. Find also
(a) distance covered in half the time of descent and
(b) time taken to cover half the distance

Explanation

When a body slides on an inclined plane, component of weight along the plane produces acceleration
`a=(mg sin theta)/m=g sin theta ="constant"`
So, equations of motion can be applied, From 2 nd equation of motion, here
`s=0+1/2(g sin theta)t^(2) [as u=0]` ....(i)
i.e, `t=sqrt((2s)/(g sin theta))=1/sin theta sqrt((2h)/g) ["as "s=h/(sin theta)]`.....(ii)
Whiel from 3 rd equation of motion,
`v^(2)=0^(2)+2(g sin theta)s`
or `v=sqrt(2gs(sin theta))=sqrt(2gh) ["as "ssintheta=h]` ...(iii)
Eqns (ii) and (iii) are the desired result. Further as from Eqn, (i) `s prop t^(2)` , so
(a) if `t'=t/2 (s')/s=((t//2)/t)^(2),"i.e " s' = s/4`
i.e, in half the time of descent the body covers only one fourth of the total distance
(b) if`s'=s/(2')((s//2)/s)=((t')/t)^(2), "i.e "t'=t/sqrt(2) ~=0.7t`
i.e, in about `0.7` times the time of descent the body covers half of the total distance.