A body starting from rest slides on an inclined plane of length s as shown in figure. Calculate the time descent and speed at the bottom. Find also distance covered in half the time of descent.

Explanation

When a body slides on an inclined plane, component of weight along the plane produces acceleration
`a=(mgsintheta)/m=gsintheta`=constant
So, equation of motion can be applied. From 2nd equation of motion, here
`s=0+1/2(gsintheta)t^2` [as u=0] ........ (i)
i.e., `t=sqrt((2s)/(gsintheta))=1/(sintheta)sqrt((2h)/g)` [as `s=h/(sintheta)`] ..........(ii)
While from 3rd equation of motion,
i.e., `v^(2)-0^2+2(gsintheta)s`
or `v=sqrt(2gs(sintheta))=sqrt(2gh)` [as `s sintheta=h`] .......(iii)
Eqns. (ii) and (iii) are the desired results . Furtheras from Eqn. (i) `spropt^2`, so
If `t.=1/2,(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 distanct.