Answer to Question #95901 in Mechanics | Relativity for Sarah

Question

A ball is thrown horizontally from the top of
a building 120 m high. The ball strikes the
ground 63 m horizontally from the point of
release. What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.

Answer 1

Let the height of the building be $H = 120m$ and the horizontal travel distance $L = 63 m$.

The equations of motion are as follows:

$v_x(t) = v_{0x}$, $v_y(t) = v_{0y} - g t = - g t$ (no initial vertical component of velocity),

$H = \frac{g t^2}{2}$, $L = v_{0x} t$ .

From the last two equations, $t =\sqrt{\frac{2 H}{g}}$, $v_{0x} = \frac{L}{t} = \frac{L \sqrt{g}}{\sqrt{2 H}}$. The speed before hitting the ground is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_{0x}^2 + (g t)^2} = \sqrt{\frac{g L^2}{2 H} + 2 g H} \approx 50.2 \frac{m}{s}$.