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(B) The initial velocity vector $\vec{u}$ makes an angle of $60^{\circ}$ with the vertical,which means it makes an angle of $30^{\circ}$ with the hori

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$20$

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(B) The initial velocity vector $\vec{u}$ makes an angle of $60^{\circ}$ with the vertical,which means it makes an angle of $30^{\circ}$ with the horizontal.$\vec{u} = 100 \cos 30^{\circ} \hat{i} + 100 \sin 30^{\circ} \hat{j} = 50\sqrt{3} \hat{i} + 50 \hat{j} \ m/s$.The velocity at time $t$ is given by $\vec{v} = \vec{u} + \vec{g}t = (50\sqrt{3} \hat{i} + 50 \hat{j}) - 10t \hat{j} = 50\sqrt{3} \hat{i} + (50 - 10t) \hat{j}$.Since the particle moves perpendicular to its initial direction,the dot product $\vec{u} \cdot \vec{v} = 0$.$(50\sqrt{3})(50\sqrt{3}) + 50(50 - 10t) = 0$.$7500 + 2500 - 500t = 0$.$10000 = 500t$.$t = 20 	ext{ seconds}$.

$A$ particle is projected upwards with a velocity of $100 \ m/s$ at an angle of $60^{\circ}$ with the vertical. Find the time when the particle will move perpendicular to its initial direction,$.....$ seconds (take $g = 10 \ m/s^2$):-

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