Two balls are thrown simultaneously from ground with same velocity of 10 ms^-1 but different angles of projection with horizontal. Both balls fall at same distance 5 $\sqrt{3}$ m from point of projection. What is the time interval between balls striking the ground?

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$\sqrt{3} s$

1 s

$(\sqrt{3} + 1) s$

$(\sqrt{3} - 1) s$

answer is A.

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Detailed Solution

Given, $5 \sqrt{3} = \frac{(10)^{2} \sin 2 θ}{g} or \sin 2 θ = \frac{\sqrt{3}}{2}$

$∴ 2 θ = 60^{0} or θ = 30^{0}$

Two different angles of projection are therefore $θ$ and 90° - 0 or 30° and 60°.

$∴ T_{1} = \frac{2 u \sin 30 °}{g} = 1 s$

$∴ T_{2} = \frac{2 u \sin 60 °}{g} = \sqrt{3} s$

$Hence, Δ t = T_{2} - T_{1} = (\sqrt{3} - 1) s$

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