If a stone is to hit at a point which is at a distance `d` away and at a height `h` (Fig. 5.200) above the point from where the stone starts, then what is the value of initial speed `u` if the stone is launched at an angle `theta` ?
.

Options

A. `(g)/(cos theta) sqrt((d)/(2(d tan theta - h)))`
B. `(d)/(cos theta) sqrt((g)/(2(d tan theta - h)))`
C. `sqrt((g d^2)/(h cos^2 theta))`
D. `sqrt((g d^2)/(d - h))`

Answer

B

Explanation

(b) `h = (u sin theta) t - (1)/(2) "gt"^2`
`d = (u cos theta) t` or `t = (d)/(u cos theta)`
`h = u sin theta (d)/(u cos theta) -(1)/(2) (g)(d^2)/(u^2 cos^2 theta)`
`u = (d)/(cos theta) sqrt((g)/(2(d tan theta - h)))`.