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At what angle with the horizontal should a ball be thrown so that the range R is related to the time of flight as R = 5 $T^{2}$ ? (Take g= 10 m $s^{- 2}$ )

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30°

45°

60°

90°

answer is B.

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Detailed Solution

$\frac{R}{T^{2}} = g \frac{\sin 2 θ}{4 \sin^{2} θ} = \frac{g}{2} \cot θ = 5 \cot θ$

Given $\frac{R}{T^{2}}$ =5, Hence, 5 = 5 cot $θ$ or $θ$ = 45°

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