Question

A body is vertically upwards. If `t_(1)` and `t_(2)` are the times at which it is height h above the point of projection while ascending and descending, respectively, find (a) the velocity of projection and height h, (b) the maximum height reached by the body and (c ) the velocity of the body at height `h//2`.

Answer 1

Time from `O` to `A` `t_(1)`
Time from `O` to `A` via `B` `t_(2)`
After `t_(1)` and `t_(2)`, the ball is above the origin at displacement `h`.

`O` to `A`: `h=ut_(1)-1/2g t_(1)^(2)`
`O` to `A` via `B: h=ut_(2)-1/2g t_(2)^(2)`
Equating (i) and (ii), we get
`ut_(1)-1/2g t_(1)^(2)=ut_(2)-1/2g t_(2)^(2)`
`u(t_(2)-t_(1))=1/2g(t_(2)^(2)-t_(1)^(2))`
`=1/2g(t_(2)+t_(1))(t_(2)-t_(1))`
`u=g/2(t_(1)+t_(2))`
`h=ut_(1)-1/2g t_(1)^(2)`
`=g/2(t_(1)+t_(2))t_(1)-1/2 g t_(1)^(2)`
`=1/2 g t_(1)t_(2)`
OR
`h=ut-1/2g t^(2)`
`g t^(2)-2ut+2h=0`
This equation is quadratic in t, it will give two value of t.
Sum of roots, `t_(1)+t_(2)=-((-2u))/g`
`implies u=g/2(t_(1)+t_(2))`
Product of roots, `t_(1) t_(2)=(2h)/g`
`impliesh=1/2g t_(1)t_(2)`
(b) Let the maximum height be `H_(max)`
`v^(2)=u^(2)-2g h_(1)`
`0=[g/2(t_(1)+t_(2))]^(2)-2g H_(max)`
`implies H_(max)=g/8(t_(1)+t_(2))^(2)`
(c ) Velocity at height `h//2`:
` v^(2)=u^(2)-2g h_(1)=u^(2)-2gh/2`
`=[g/2(t_(1)+t_(2))]^(2)-g. 1/2 g t_(1)t_(2)`
`=g^(2)/4(t_(1)^(2)+t_(2)^(2)+2t_(1)t_(2))-g^(2)/2t_(1)t_(2)`
`=g^(2)/4(t_(1)^(2)+t_(2)^(2))`
`v=g/2(t_(1)^(2)+t_(2)^(2))^(1//2)`