A particle is projected with `10ms^(-1)` at angle `theta` above horizontal from a horizontal ground. Find the value of `theta` so that area under the parabolic path of the projectile will be maximum
`(g=10ms^(-2))`.

Options

A. `60^(@)`
B. `45^(@)`
C. `90^(@)`
D. None of these

Explanation

`A=int y dx=int_(0)^(R)tan theta(x-(x^(2))/(R))dx`
`=tan theta((R^(2))/(2)-(R^(3))/(3R))`
`(dA)/(dtheta)=0=2 sin^(2)thetacos^(2)theta-sin^(3)thetasintheta=0`
`=tan theta(R^(2))/(6)=(tantheta)/(6)((u^(2)sin theta cos theta)/(g))^(2)`
`sin^(2)theta(3 cos^(2)theta-sin^(2)theta)=0`
`=(10^(4)xxu^(2))/(36xx10xx10cos theta)sin^(2)thetacos^(2)theta`
`3=tan^(2)theta`
`=(200)/(3)sin^(3)theta cos theta`
`tan theta=sqrt(3)`
`theta=60^(@)`