From the top of a tower, two balls are thrown horizontally with velocities u1 and u2 in opposite directions. If their velocities are perpendicular…

Q: From the top of a tower, two balls are thrown horizontally with velocities u1 and u2 in opposite directions. If their velocities are perpendicular to each other just before they strike the ground, find the height or tower.

Sol: Let h = height of the tower

$\large t = \sqrt{\frac{2 h}{g}}$

At time of reaching ground velocities are $\vec{v_1} = u_1 \hat{i} – g t\hat{j}$ and $\vec{v_2} = – u_2 \hat{i} – g t\hat{j}$

As velocities are perpendicular to each other hence ,

$\large \vec{v_1} . \vec{v_2} = 0 $

$\large (u_1 \hat{i} – g t\hat{j}) . (- u_2 \hat{i} – g t\hat{j}) = 0 $

$\large u_1 u_2 = g^2 t^2 $

$\large t = \sqrt{\frac{u_1 u_2}{g^2}}$

$\large \sqrt{\frac{2 h}{g}} = \sqrt{\frac{u_1 u_2}{g^2}}$

$\large h = \frac{u_1 u_2}{g}$