A particle is projected from the ground at angle such that `t_(1)` time it just clears the top of a pole in its path. It takes further `t_(2)` time to reach the ground. What is height of the pole?

Explanation

Height of the building is equal to the vertical displacement of the particle at time `t_1`
Vertical displacement `y= ut_1 + (1)/(2) a t_(1)^(2) = ut_1 - (1)/(2) "gt"_(1)^(2)………….` (i)
and total flight time `t_1 + t_2 = ( 2u_2)/(g) rArr u_1 = (g)/(2) (t_1 +t_2)`
put value `u_2` in equation (i) `y = (g)/(2)(t_1 + t_2) t_1 - (1)/(2) g(t_1)^(2) = (1)/(2)"gt"_1 t_2`. so height of the building `= (1)/(2) "gt"_1t_2`.