Show that there are two values of time for which a projectile is at the same height. Also show mathematically that the sum of these two times is equal to the time of flight.

Explanation

Refer to fig. 2 (d) . 31, let the projectile projected from (O) reach the locitions ` A and B` after times `t_(1) and t_(2)`.
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If the projectile reaches at a height (h) after time (t), then ` u=u sin theta, a=- g, S=h, t=t`
` S=ut + 1/2 at&(2)`
:. h=u sing theta t + 1/2 9-g) t^(2)`
`gt^(2) -2 u sin theta t + 2 h=0`
or ` t= (2 u sin thea +- sqrt (4 u sin^(2) theta -4 g xx 2 h)/(2 g)
:. ` t=(u sin theta -sqrt(u^(2) sin^(2) theta -4 g xx 2 h))/ (2 g)`
and `t_(1) =( u sin theta + ssqrt (u^(2) sin^(2) sin^(2) theta -2 gh))/g`
` + (u sin theta + sqrt (u^(2) sin ^(2) theta-2 gh))/g`
`= (2 u sin theta)/g =T =time of flght.