Answer to Question #94513 in Mechanics | Relativity for Evan Meinhold

Question

A small rock is thrown vertically upward with a speed of 27.0 m/sfrom the edge of the roof of a 31.0-m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. Ignore air resistance. What is the speed of the rock just before it hits the street? How much time elapses from when the rock is thrown until it hits the street?

Answer 1

Let us split the motion into two parts: first part - when the rock is moving upwards until it stops, and the second - when the rock is accelerating down from the maximum height with zero initial speed.

Equations of motion for the first part are:

$y(t) = H + v_0 t - \frac{g t^2}{2}$ , $v(t) = v_0 - gt$ , where $v_0$ is the initial upward speed and $H$ is the height of the building.

When the rock reaches the highest point, it stops, hence $v(t') = v_0 - g t' = 0$, from where the time it takes to reach the highest point is $t' = \frac{v_0}{g}$ . The maximum height is then $h_{max} = y(t') = H + \frac{v_0^2}{g} - \frac{g}{2}\frac{v_0^2}{g^2} = H + \frac{v_0^2}{2 g}$ .

Equations of motion for the second part are:

$y(t) = h_{max} - \frac{g t^2} {2}$ , $v(t) = g t$ (start falling down from the highest point with initial zero speed).

When the rock reaches the ground $y(t'') = h_{max} - \frac{g {t''}^2}{2} = 0$ , from where $t'' = \sqrt{\frac{2 h_{max}}{g}}$ .

The total time it takes to reach the ground is $T = t' + t'' = \frac{v_0}{g} + \sqrt{\frac{2 h_{max}}{g}} = \frac{v_0}{g} +\sqrt{\frac{2(H+\frac{v_0^2}{2 g})}{g}}$ , and the speed when the rock hits the ground is $v = v(t'') = g t'' = \sqrt{2 g h_{max}} = \sqrt{2 g H + v_0^2}$ .

Substituting $H = 31 m$ , $v_0 = 27 \frac{m}{s}$ , obtain $T = 6.48 s$, $v = 36.57 \frac{m}{s}$.