A small disc A slides down with initial velocity equal to zero from the top of a smooth hill of height H having a horizontal portion. What must be the height of the horizontal portion h to ensure the maximum distance s covered by the disc? What is it equal to?

Explanation

Applying work energy theorem for motion from A to B

`mg(H-h)=(1)/(2)m(V^(2)-0)`
`therefore V=sqrt(2g(H-h))`
If time taken by the disc to move from point B to the point C on ground is t, then
`Deltay=u_(y)t-(1)/(2)"gt"^(2)=h`
`therefore0-(1)/(2)"gt"^(2)=-himpliest=sqrt((2h)/(g))`
`S=Deltax=vt`
`therefore S=sqrt(2g(H-h)xx(2h)/(g))=2sqrt(Hh-h^(2)).....(i)`
for s to be maximum, `(Hh-h^(2))` is maximum
i.e. `(d)/(dh)[Hh-h^(2)]=0 therefore H-2h=0`
`impliesh=(H)/(2)`
Putting this value of h in equation (i)
`therefore S_(max)=2sqrt(H((H)/(2))-((H)/(2))^(2))=H`