Question

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, the angle between velocities of balls becomes ${90}^o$?

• A. $\frac{2\sqrt{v_1{v}_2}}{g}$
• B. $\frac{\sqrt{v_1{v}_2}}{g}$
• C. $\frac{g}{\sqrt{v_1{v}_2}}$
• D. $\frac{\frac{g}{\sqrt{v_1{v}_2}}}{2g}$

Answer 1

Answer: (B)

$\frac{\sqrt{v_1{v}_2}}{g}$

General velocity vector of the projectiles is given by

$\overrightarrow{V}=v_1+j\left(gt\right)$...................(1)

$\overrightarrow{V}=v_2+j\left(gt\right)$.............(2)

for both the vectors above to be perpendicular ,their dot product must be zero.

hence,dot product is

$v_1\times v_2+g^2{t}^2=0$

finally,

$t=\frac{\sqrt{v_1{v}_2}}{g}$