A ball is thrown from a point in level with velocity `u` and at a horizontal distance `r` from the top os a tower of height `h`.
How must the speed and angle of the projection of the ball be related to `r` in order that the ball may just go grazing the top edge of the tower ?

Options

A. `gr = u^2 sin 2 theta`
B. `gr = u^2 sin theta`
C. `gr = u^2 cos 2 theta`
D. `gr = u^2 cos theta`

Answer

C

Explanation

( c) Here `r` will become range
`r = (u^2 sin 2 theta)/(g)`
`rArr gr = u^2 sin 2 theta`
`h = u sin theta t + (1)/(2) "gt"^2`
`x = u cos theta t`
`t = (x)/(u cos theta)`
`h = u sin theta (x)/(u cos theta) + (1)/(2) g (x^2)/(u^2 cos^2 theta)`
`gx^2 + 2u^2 sin theta cos theta x - 2h u^2 cos^2 theta = 0`
`x = (u cos theta)/(g)[sqrt(u^2 sin^2 theta + 2gh - u sin theta)]`.
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