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_1^2 + h_2t_2^2)/(h_2t_1 + h_1t_2)`
C. `(h_1t_2^2 + h_2t_1^2)/(h_1 t_2 + h_2t_1)`
D. `(h_1 t_1^2 - h_2 t_2^2)/(h_1t_1 - h_2t_2)`

Answer

A

Explanation

Let a cricket ball be thrown with velocity u at an angle `theta` with the horizontal .
As per question, `h_(1) = u sintheta t_(1) -1/2 "gt"_(1)^2`
or `u sin theta t_(1) = h_(1) +1/2 "gt"_(1)^(2) " "...(i)`
and `h_(2) = u sin theta t_(2) -1/2 "gt "_(2)^(2)`
or `u sin theta t_(2) = h_(2) +1/2"gt"_(2)^(2) `
Divide eqn. (i) by eqn. (ii) , we get
`t_1/t_2 =(h_1+1/2"gt"_(1)^2)/(h_2+1/2"gt"_2^2),h_(2)t_(1)+1/2"gt"_(2)^(2)t_(1)=h_1t_(2)+1/2"gt"_(1)^(2)t_(2)`
`h_(1)t_(2) -h_(2)t_(1) =1/2 g(t_1t_2^2-t_1^2t_(2))" "...(iii)`
Time of fight , `T=(2u sin theta)/g =2/g[(h_1+1/2"gt"_1^2)/t_1]` (Using (i))
`=2/gh_1/t_1+t_1=h_1/t_1((t_1t_2^2-t_1^2t_2)/(h_1t_2-h_2t_1))+t_1" "` (Using (iii))
`=(h_1t_2^2-h_1t_1t_2)/(h_1h_2-h_2h_1)+t_1=(h_1t_2^2-h_2t_1^2)/(h_1t_2-h_2t_1)`