A particle is projected vertically upwards from a point O on the ground. It takes time `t_(1)` to reach a point `A` at a height h above the ground, it continues to move and takes a time `t_(2)` to reach the ground. Find (a) h, (b) the maximum height reached and (c ) the velocity of the partical at the half of maximum height.

Explanation

Time from `O` to `A` `t_(1)`
Time from `A` to `O` via `B`: `t_(2)`
`O` to `O: t=t_(1)+t_(2)`
`h=ut-1/2g t^(2)`
`0=u(t_(1)+t_(2))-1/2g (t_(1)+t_(2))^(2)`
`u=(g(t_(1)+t_(2)))/2`
`O` to `A: h=ut_(1)-1/2 g t_(1)^(2)=1/2 g t_(1)t_(2)`
`O` to `B: v^(2)=u^(2)-2g h_(1)`
`0=u^(2)-2gH_(max)`
`H_(max)=u^(2)/(2g)=g/8(t_(1)+t_(2))^(2)`
Velocity at half of the maximum height
`v_(1)^(2)=u^(2)-2gH_(max)/2=g^(2)/4(t_(1)+t_(2))^(2)-g^(2)/8(t_(1)+t_(2))^(2)`
`=g^(2)/8(t_(1)+t_(2))^(2)`
`v_(1)=g/(2sqrt(2))(t_(1)+t_(2))`