Question

A cricket ball thrown across a field is at heights h₁ and h₂ from the point of projection at times t₁ and t₂ 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

पर्याय

• `(h_1t_1^2 - h_2t_1^2)/(h_1t_2 - h_2t_1)`

• `(h_1t_1^2 + h_2t_2^2)/(h_2t_1 - h_1t_2)`

• `(h_1t_2^2 + h_2t_1^2)/(h_1t_2 + h_2t_1)`

• `(h_1t_1^2 + h_2t_2^2)/(h_1t_1 + h_2t_2)`

Answer 1

`(h_1t_1^2 - h_2t_1^2)/(h_1t_2 - h_2t_1)`

Explanation:

`h_1 = (u sin theta)t_1 - 1/2 "gt"_1^2` ......(i)

`h_2 = (u sin theta)t_2 - 1/2 "gt"_2^2` ......(ii)

∴ `((h_1 + 1/2 "gt"_1^2))/((h_2 + 1/2 "gt"_2^2)) = t_1/t_2`

∴ `h_1t_2 - h_2t_1 = g/2(t_1t_2^2 - t_1^2t_2)` ......(iii)

∴ T = `(2 u sin theta)/g`

= `2/g[(h_1 + 1/2 "gt"_1^2)/t_1]`  ......[From (i)]

= `2/t_1 [h_1/g + t_1^2/2]`

= `h_1/t_1 xx ((t_1t_2^2 - t_1^2 t_2)/(h_1t_2 - h_2t_1)) + t_1`  ......[From (iii)]

= `(h_1t_1^2 - h_2t_1^2)/(h_1t_2 - h_2t_1)`