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` ?
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Options

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

Answer

B

Explanation

` h= (u sin theta ) t - 1/2 gt^2`
` d= u cos hteta t ` or ` t = d//(u cos theta)`
From (i), ` h= u sin theta xx d/( u cos theta) - 1/2 g d^2 /(u^2 cos^2 theta0`
or ` u = d/(cos theta) sqrt (g/( 2 (d tan theta - h))`.