A projectile is fired with a velocity u at right angles to the slope, which is inclined at an angle `theta` with the horizontal. Derive an expression for the distance R to the point of impact.
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Answer

A, B, C

Explanation

`x = ( u sin theta) T`
`y = (1/2 gT^2) - (u cos theta)T`
`y/x = tan theta = (1/2 gT^2 - (u cos theta) T)/((u sin theta)T)`
`:. (2u sin theta) (tan theta) = gT - (2u cos theta)`
`T = (2u[sin theta tan theta + cos theta])/g`
`=(2usectheta)/g`

Now, `R = (x) sec theta`
`=(u sin theta T) sec theta`
`=(u sin theta)((2u sec theta)/g) (sec theta)`
`=(2u^2sin theta sec^2theta)/g`.