Question

A ball is thrown horizontally at a velocity of 10.0 m/s from the top of a 90.0 m building. Calculate the distance from the base of the building that the ball is safely caught on the ground.

Answer 1

Horizontal velocity will be 10 m/s, initial vertical velocity will be 0.

Now, For horizontal motion, at any time t,

$x = vt$

For vertical motion,

$y = \frac{1}{2}gt^2$

From both equations, removing t

we get,

$y = \frac{1}{2}g (\frac{x}{v})^2 = \frac{g}{2v^2}x^2$

$\implies x = \sqrt{\frac {2 v^2 y}{g}} = (\sqrt{\frac {2 y}{g}})v$

Now putting the values,

$x = (\sqrt{\frac {2 *90}{10}})10 = 42.43 m$