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(D) Let the velocity of the projectile at point $P$ be $v$. Since the velocity at $P$ is perpendicular to the inclined plane,the component of velocity

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$Tv 	an 	heta$

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(D) Let the velocity of the projectile at point $P$ be $v$. Since the velocity at $P$ is perpendicular to the inclined plane,the component of velocity parallel to the inclined plane at $P$ is zero.Let $u$ be the initial velocity at $Q$ and $\alpha$ be the angle of projection with the inclined plane. The component of acceleration parallel to the inclined plane is $-g \sin 	heta$.Using the equation of motion along the inclined plane: $v_p = u_p + a_p T$,where $v_p = 0$ (velocity parallel to the plane at $P$ is zero).$0 = u \cos \alpha - g \sin 	heta T \implies u \cos \alpha = g \sin 	heta T$.The distance $PQ$ is the range along the inclined plane,given by $PQ = (u \cos \alpha) T - \frac{1}{2} (g \sin 	heta) T^2$.Substituting $u \cos \alpha = g \sin 	heta T$ into the equation for $PQ$:$PQ = (g \sin 	heta T) T - \frac{1}{2} g \sin 	heta T^2 = \frac{1}{2} g \sin 	heta T^2$.Also,from the condition that velocity at $P$ is perpendicular to the plane,the time of flight $T$ to reach $P$ is $T = \frac{u \cos \alpha}{g \sin 	heta}$.Given the velocity at $P$ is $v$,and it is perpendicular to the plane,$v$ is the component of velocity perpendicular to the plane at $P$. $v = u \sin \alpha - g \cos 	heta T$.Since $v$ is the final velocity at $P$ and it is perpendicular to the plane,the horizontal component of velocity is $v \sin 	heta = u \cos \alpha$.Thus,$PQ = (v \sin 	heta) T / \cos 	heta = v T 	an 	heta$.

If the time taken by the projectile to reach point $P$ from point $Q$ is $T$,and the velocity at $P$ is perpendicular to the inclined plane,then $PQ$ =

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