A cricket ball thrown across a field is at heights `h_(1)` and `h_(2)` from the point of projection at time `t_(1)` and `t_(2)` respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

Options

A. `(h_1t_2^2 -h_2t_1^2)/(h_1t_2-h_2t_1)`
B. `(h_1t_2^2 +h_2t_1^2)/(h_1t_2+h_2t_1)`
C. `(h_1t_2)/(h_2t_1-h_1t_2)`
D. None

Answer

A

Explanation

`h_1=u sin theta t_1 + 1/2 g t_1^2 , h_2 = u sin theta t_2 + 1/2 g t_2^2`
So, `t_1/t_2=(h_1+1/2 g t_1^2)/(h_2+1/2 g t_2^2) rArr h_1t_2-h_2t_1=1/2 g (t_1t_2^2-t_1^2t_2)`
Time of flight = `2u sin theta//g =(h_1t_2^2 -h_2t_1^2)/(h_1t_2-h_2t_1)` [ Use above eqn to simplify ]