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Standard XII

Physics

Average Velocity in 2D Motion

There are thr...

Question

There are three paths a, b and c of a projectile projected from point $P$ as shown in Fig. Prove that $v_{1} > v_{2}$ and $v_{3} = v_{4}$ . Which path is correct?

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Solution

Let $α$ be the angle of projection with incline and $β$ be the angle at which projectile strikes the plane.
Then $tan α = v_{3} / v_{1}$ and $tan β = v_{4} / v_{2}$ . Here the magnitude of velocity component initially and finally along $y - a x i s$ will be same because $S_{y} = 0$ Hence $v_{3} = v_{4}$ .But velocity component along $x - a x i s$ goes on decreasing .So $v_{2} < v_{1}$ . Hence from above equations, $tan β > tan α \Rightarrow β > α$ .Hence appropriate path should be $b$ .

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There are three paths a, b and c of a projectile projected from point P as shown in Fig. Prove that v1>v2 and v3=v4. Which path is correct?

There are three paths a, b and c of a projectile projected from point $P$ as shown in Fig. Prove that $v_{1} > v_{2}$ and $v_{3} = v_{4}$ . Which path is correct?