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JEE PHYSICS MCQ: MOTION IN PLANE

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A ball is thrown at an angle θ\theta and another ball is thrown at an angle (90∘−θ)\left(90^{\circ}-\theta\right) with the horizontal direction from the same point with velocity 39.2 ms−139.2 \mathrm{~ms}^{-1}. The second ball reaches 50 m50 \mathrm{~m} higher than the first ball. Find their individual heights. Take g=9.8 ms−2{g}=9.8 \mathrm{~ms}^{-2}.
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Correct option: A
14.2 m,64.2 m14.2 \mathrm{~m}, 64.2 \mathrm{~m}
Option A (14.2 m,64.2 m14.2 \mathrm{~m}, 64.2 \mathrm{~m}) is correct because the individual heights are calculated using projectile motion equations and the given condition that the second ball reaches 50m higher than the first.

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