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(A) Let $u$ be the initial velocity and $	heta$ be the angle of projection.The initial velocity vector is $\vec{v}_1 = u \cos 	heta \hat{i} + u \sin

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Let $u$ be the initial velocity and $	heta$ be the angle of projection.The initial velocity vector is $\vec{v}_1 = u \cos 	heta \hat{i} + u \sin 	heta \hat{j}$.After time $t$,the velocity vector is $\vec{v}_2 = u \cos 	heta \hat{i} + (u \sin 	heta - gt) \hat{j}$.Since the direction of motion is perpendicular to the initial direction,the dot product of the two velocity vectors must be zero: $\vec{v}_1 \cdot \vec{v}_2 = 0$.$(u \cos 	heta \hat{i} + u \sin 	heta \hat{j}) \cdot (u \cos 	heta \hat{i} + (u \sin 	heta - gt) \hat{j}) = 0$$u^2 \cos^2 	heta + u \sin 	heta (u \sin 	heta - gt) = 0$$u^2 \cos^2 	heta + u^2 \sin^2 	heta - ugt \sin 	heta = 0$$u^2 (\cos^2 	heta + \sin^2 	heta) = ugt \sin 	heta$$u^2 = ugt \sin 	heta$$t = \frac{u}{g \sin 	heta}$Given $u = 15 \ m/s$,$g = 10 \ m/s^2$,and $	heta = 30^{\circ}$:$t = \frac{15}{10 	imes \sin 30^{\circ}} = \frac{15}{10 	imes 0.5} = \frac{15}{5} = 3 \ s$.

$A$ projectile is thrown into air with velocity $15 \ m/s$ at an angle $30^{\circ}$ with the horizontal. After what time is its direction of motion perpendicular to its initial direction (in $s$)? (Assume $g = 10 \ m/s^2$)

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