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(B) Let the initial velocity be $u$ at an angle $\alpha$ with the horizontal. The horizontal component is $u_x = u \cos \alpha$ and the vertical compo

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$\frac{u}{g \sin \alpha}$

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(B) Let the initial velocity be $u$ at an angle $\alpha$ with the horizontal. The horizontal component is $u_x = u \cos \alpha$ and the vertical component is $u_y = u \sin \alpha$.After time $t$,let the velocity be $v$. The horizontal component remains constant: $v_x = u \cos \alpha$.The vertical component becomes $v_y = u \sin \alpha - gt$.For the velocity to be at right angles to the initial direction,the dot product of the initial velocity vector $\vec{u}$ and the final velocity vector $\vec{v}$ must be zero,or more simply,the slope of the new velocity must be the negative reciprocal of the initial slope.Alternatively,using the condition that the velocity vector $\vec{v}$ is perpendicular to $\vec{u}$,we have $\vec{u} \cdot \vec{v} = 0$.$(u \cos \alpha \hat{i} + u \sin \alpha \hat{j}) \cdot (u \cos \alpha \hat{i} + (u \sin \alpha - gt) \hat{j}) = 0$$u^2 \cos^2 \alpha + u \sin \alpha (u \sin \alpha - gt) = 0$$u^2 \cos^2 \alpha + u^2 \sin^2 \alpha - ugt \sin \alpha = 0$$u^2 (\cos^2 \alpha + \sin^2 \alpha) = ugt \sin \alpha$$u^2 = ugt \sin \alpha$$t = \frac{u}{g \sin \alpha}$

If at any point on the path of a projectile its velocity is $u$ at an inclination $\alpha$ to the horizontal,then after what time will it move at right angles to its former direction?

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