A ball is projected vertically up such that it passes thorugh a fixes point after a time `t_(0)` and `t_(2)` respectively. Find
a. The height at which the point is located with respect to the point of projeciton
b. The speed of projection of the ball.
c. The velocity the ball at the time of passing through point `P`.
d. (i) The maximum height reached by the balll relative to the point of projection `A` (ii) maximum height reached by the ball relative to point `P` under consideration.
e. The average speed and average velocity of the ball during the motion from `A` to `P` for the time `t_(1)` and `t_(2)` respectively.
.

Explanation

Let the ball be projected up with an initial velocity `u`,
It passes through point `P` at `t=t_(1)` during its ascent and at
`t=t_(2)` during its descent.
For the motion of the ball from `A` to `P`,
`s=+h, v_(0)=u, a=-g`, and `t=t_(1)` and `t_(2)`
Substituting the above value, in `s=v_(0)t+(1)/(2)a t^(2)`, we get
`{:(h=ut_(1)-(1)/(2) g t_(1)^(2) ("upward motion")),(=ut_(2)-(1)/(2)g t_(2_(2)^(2)) ("downward motion")):}]`
Hence, from (i) `ut_(1)-(1)/(2)g t_(1)-(1)/(2)g t^(2)=ut_(2)-(1)/(2)g t_(2)^(2)`
`u(t_(2)-t_(1))=(1)/(1)g(t_(2)-t_(1))(t_(2)+t_(1)) rArr u(1)/(2)g (t_(2)+t_(1)`
Subsstituting the value of `u` in `i` we get `h=(1)/(2)g t_(1)t_(2)`
Solving the above equation, we have
`h=(g t_(1)t_(2))/(2)` and `v=(g(t_(1)+t_(2)))/(2)`
c. Using relation `vec v=vec u+vecat+vec at`, we get
`v_(p)=((g(t_(2)+t_(1)))/(2))-g t_(1)=(g)/(2)(t_(2)-t_(1))`
d. Maximum height reached by ball from the point of projection.
Using `v^(2)=u^(2)+2` as
`0=((g)/(2)(t_(2)+t_(1)))^(2) -2gH rArr H_(max)=(g)/(8)(t+(1)+1_(2))^(2)`
ii. Height reached by ball from point `P`,
`h' =H_(max)-h=(g)/(8)(t_(1)+t_(2))^(2)-(1)/(2)g `t_(1)t_(2) =(g)/(8) (t_(2)-t_(1))^(2)`
e.i. `A` to `P` for time `t_(1)`
The average speed and average veloctty from `A` to `P` willbe same as the particle is moving in straight line without changing the direction.
`=(Delta)/(Deltat)=(h)/(t_(1))(1)/(2) g t_(1) (t_(2))/(t)=(1)/(2) g t_(2)`
ii. Motion from `A` to `P` for time `t_(2)`.
Average speed `=(Total distance)/(Total time)`
`=(h+2h')/(t_(2))`
`((1)/(2)g t_(1)t_(2)+2((g)/(8)(t_(2)-t_(1)^(2))))/(t_(2))`
`(g(t_(1)^(2)+t_(2)^(2)))/(4t_(2))`
Motion from `A` to `P` for time `t_(1)`,
Average velocity `lt vec vg t =(Delta vecy)/(Deltavect)=(h)/(t_(2))=((1)/(2)g t_(1)t_(2))/(4t_(2))`.