If a stone is to hit at a point which is at a distance d away and at a height h above the point from where the stone starts as shown in the figure bo then what is the value of initial speed u if stone is launched at an angle `theta` ?

Options

A. `g/(costheta)sqrt((d)/(2(dtan theta-h)))`
B. `d/(costheta)sqrt((g)/(2(dtan theta-h)))`
C. `sqrt((gd^2)/(hcos^(2) theta))`
D. `sqrt((gd^2)/((d-h)))`

Answer

B

Explanation

Motion along horizontal direction
`d = ucos theta xx t " or " t = (d)/((u cos theta) ) " "...(i)`
Motion along vertical direction
`h = ( u sin theta) t = -1/2 "gt"^(2) " "...(ii)`
Substitute the value of t in equation (ii), we get
`h = u sin theta xx (d)/(u cos theta) -1/2 g (d^2)/(u^2 cos ^(2) theta)`
or `h = d tan theta -1/2 g (d^2)/(u^(2) cos ^(2) theta)`
or `1/2 (gd^2)/(u^2 cos^(2) theta) = d tan theta- h " or " u^(2) = (gd^2)/(2(dtantheta - h ) cos^(2) theta)`
or `u = d/(cos theta) sqrt((g)/(2(dtan theta-h)))`