A ball is projected with a velocity $V_{0}$ at angle $θ$ with horizontal. Then its velocity is normal to the initial velocity of projection at time:

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$t = \frac{V_{0}}{g \sin θ} for 0 < θ \leq \frac{π}{2}$

$t = \frac{V_{0}}{g \sin θ} for 0 < θ \leq \frac{π}{4}$

$t = \frac{V_{0}}{g \sin θ} for \frac{π}{4} < θ \leq \frac{π}{2}$

$t = \frac{V_{0}}{g \sin θ} for all value of θ$

answer is C.

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Detailed Solution

Given
Velocity= $V_{0}$

$\begin{array}{l} I n i t i a l l y, {\vec{V}}_{0} = V_{0} cosθ \hat{i} + V_{0} sinθ \hat{j} \\ After some time t, \vec{V} = V_{0} cosθ \hat{i} + (V_{0} sinθ- g t) \hat{j} \\ If {\vec{V}}_{0} \cdot \vec{V} = 0 then t = \frac{V_{0}}{g \sin θ} \\ T i m e o f f l i g h t T = \frac{2 V_{0} \sin θ}{g} \\ G i v e n, t < T \\ \frac{V_{0}}{g \sin θ} \leq \frac{2 V_{0} \sin θ}{g} \\ \Rightarrow \sin θ \geq \frac{1}{\sqrt{2}} \\ ∴ \frac{π}{4} \leq θ \leq \frac{π}{2} \end{array}$