A cricket ball thrown across a field is a 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_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2) -h_(2)t_(1)))`
B. `((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2)-h_(2)t_(1)))`
C. `((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2)-h_(2)t_(1)))`
D. None of these

Answer

C

Explanation

For vertically moment,
`h_(1) = u sin theta t_(1) -(1)/(2) g t_(1)^(2)` (for `h_(1))`
`rArr t_(1) = (h_(1)+(1)/(2)g t_(1)^(2))/(u sin theta)` ........(i)
`rArr h_(2) = u sin theta t_(2) - (1)/(2) g t_(2)^(2)` (for `h_(2))`
`rArr t_(2) = (h_(2)+(1)/(2)g t_(2)^(2))/(u sin theta)` ........(ii)
On dividing Eq. (i) by Eq. (ii), we get
`(t_(1))/(t_(2)) = (h_(1)+(1)/(2)g t_(1)^(2)//u sin theta)/(h_(2)+(1)/(2)g t_(2)^(2)//u sin theta) rArr h_(1)t_(2)-h_(2)t_(1) = (g)/(2) (t_(1)t_(2)^(2)-t_(1)^(2)t_(2))`
The time of flight of the ball,
`T = (2u sin theta)/(g) = (2)/(g) (u sin theta)` [from Eq.(i)]
`= (2)/(g) ((h_(1)+1//2g t_(1)^(2))/(t_(1))) =(2)/(t_(1)) ((h_(1))/(g)+(t_(1)^(2))/(2))`
`=(h_(1))/(t_(1)) xx (2)/(g)+t_(1) = (h_(1))/(t_(1)) xx ((t_(1)t_(2)^(2)-t_(1)^(2)t_(2))/(h_(1)t_(2)-h_(2)t_(1))) +t_(1)`
`=(h_(1)t_(1)t_(2)^(2)-h_(1)t_(1)^(2)t_(2)+h_(1)t_(1)^(2)t_(2)-h_(2)t_(1)^(3))/(t_(1)(h_(1)t_(1)h_(2)t_(1)))=((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2)-h_(2)t_(1)))`