1.

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)).

Answer»

`60^(@)`
`45^(@)`
`90^(@)`
NONE of these

Solution :`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^(@)`


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