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(B) Let the velocities of the two balls at time $t$ be $\vec{v}_A$ and $\vec{v}_B$.For the first ball thrown with velocity $v_1$ in the positive $x$-d

Vedclass · Accepted Answer

$\frac{\sqrt{v_1 v_2}}{g}$

Vedclass · Answer

(B) Let the velocities of the two balls at time $t$ be $\vec{v}_A$ and $\vec{v}_B$.For the first ball thrown with velocity $v_1$ in the positive $x$-direction: $\vec{v}_A = v_1 \hat{i} - gt \hat{j}$.For the second ball thrown with velocity $v_2$ in the negative $x$-direction: $\vec{v}_B = -v_2 \hat{i} - gt \hat{j}$.The angle between the velocities is $90^o$ when their dot product is zero: $\vec{v}_A \cdot \vec{v}_B = 0$.$(v_1 \hat{i} - gt \hat{j}) \cdot (-v_2 \hat{i} - gt \hat{j}) = 0$.$-v_1 v_2 + g^2 t^2 = 0$.$g^2 t^2 = v_1 v_2$.$t^2 = \frac{v_1 v_2}{g^2}$.$t = \frac{\sqrt{v_1 v_2}}{g}$.

Two balls are thrown horizontally from the top of a tower with velocities $v_1$ and $v_2$ in opposite directions at the same time. After how much time will the angle between the velocities of the balls become $90^o$?

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