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(D) The equation of the trajectory is given by $y = ax - bx^2$. Comparing this with the standard equation of projectile motion,$y = x 	an 	heta - \f

Vedclass · Accepted Answer

$\frac{a^2}{4b}, 	an^{-1}(a)$

Vedclass · Answer

(D) The equation of the trajectory is given by $y = ax - bx^2$. Comparing this with the standard equation of projectile motion,$y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$,we get: $a = 	an 	heta$ and $b = \frac{g}{2u^2 \cos^2 	heta}$. From $a = 	an 	heta$,the angle of projection is $	heta = 	an^{-1}(a)$. The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$. Since $	an 	heta = a$,we have $\sin 	heta = \frac{a}{\sqrt{1+a^2}}$ and $\cos 	heta = \frac{1}{\sqrt{1+a^2}}$. Also,$b = \frac{g}{2u^2 \cos^2 	heta} \implies u^2 = \frac{g}{2b \cos^2 	heta}$. Substituting $u^2$ into the formula for $H$: $H = \frac{g}{2b \cos^2 	heta} \cdot \frac{\sin^2 	heta}{2g} = \frac{	an^2 	heta}{4b} = \frac{a^2}{4b}$. Thus,the maximum height is $\frac{a^2}{4b}$ and the angle of projection is $	an^{-1}(a)$.

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