An onject is thrown vertically upward with some speed. It crosses ` 2` points ` p,q` whith are separated by (h) metre. If (t) is the time between (p) and highest point and coming back and ` t_(q)` is the time between `q` and highest point and coming back, relate acceleration due to gravity `t_(p),t_(q) and h`.

Explanation

Let (u') be the velocity of the object while crossing point ` q. Fig. 2 (b) .29.` A is the highest point of vertical motion of object . As per question, the time taken by the object in going from ` p to A= T_(p) //2.` The time taken by the object in going from (q) to ` A=t_(q)//2`.
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Taking vertical upward motion fo object from ` pto A`, we have
` u=u' , v=0, a=- g, t=t_(p)//2`
As, ` v=v +at`
:. ` v'^(2) =u'^(2) + 2 (-g) h`
or ` 2 gh =u'^(2) -v'^(2) = (g^(2) t_(p)^(2))/4 -(g^(2) t_(q)^(2))/4`
or ` g=(8 h) /((t_(p)^(2) -t_(q)^(2))`.