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`.
At what horizontal distance `x` from the foot of the tower does the ball hit the ground ?

Options

A. `(u cos theta)/(g){(u^2 sin^2 theta + gh)^(1//2) - u sin theta}`
B. `(u cos theta)/(g){(u^2 cos^2 theta + 2gh)^(1//2) - u cos theta}`
C. `(u cos theta)/(g){(u^2 cos^2 theta + gh)^(1//2) - u cos theta}`
D. `(u cos theta)/(g){(u^2 cos^2 theta + gh)^(1//2) - u cos theta}`

Answer

A

Explanation

( a) 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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