A shot is fired with a velocity `u` at a very high vertical wall whose distance from the point of projection is `x`.The greatest height above the level of the point of projection at which the bullet can hit the wall is.

Options

A. `(u^(4)+g^(2)x^(2))/(2gu^(2))`
B. `(u^(4)-g^(2)x^(2))/(gu^(2))`
C. `(u^(4)-g^(2)x^(2))/(4gu^(2))`
D. `(u^(4)-g^(2)x^(2))/(2gu^(2))`

Answer

D

Explanation

Let `theta` the angle of projection, suppose `y` is the height at which bullet hit the wall.We have, from equation of trajectory.
`y=x tan theta-(gx^(2)sec^(2)theta)/(2u^(2))`----(i)
`(dy)/(d theta)= x sec^(2) theta-(gx^(2))/(2u^(2))2sectheta(sec theta tan theta)`
`=x sec^(2) theta [1-(xg tan theta)/u^(2)]=0 or tan theta=u^(2)/(gx)`
Substituting this value in equation (i), we get
`y_(max)=xu^(2)/(gx)-1/2(gx^(2))/u^(2)[1-u^(4)/(x^(2)g^(2))]`
`=u^(2)/g-(gx^(2))/(2u)-u^(2)/(2g)=(2u^(4)-g^(2)x^(2)-u^(4))/(2u^(2)g)` or `y_(max)=(u^(4)-g^(2)x^(2))/(2gu^(2))`