Question

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?

Answer 1

Correct Answer - B
In order to obtain the velocity point (B), we apply the law of conservation of energy.
So,
Loss in `PE`=Gain in `KE`
`m g (H-h) =1/2mv^(2)`
`:. v=(sqrt[2g(H-h)])`
Further `h=1/2gt^(2)`
`:. t=sqrt((2h)/g)`
Now, `s=v xx t =sqrt[[2g(H-h)]] xx sqrt((2 h//g))`
or `s=sqrt[[4h(H-h)]]` .....(i)
For maximum value of s, `(ds)/(dh)=0`
`:. 1/(2sqrt[[4h(H-h)]]xx4(H))=0` or `h=H/2`
Substituting `h=H//2` in E.q. `(1)`, we get
`s=sqrt[[4(h//2)(H-//2)]]=sqrt(H^(2))=H`