Answer to Question #87050 in Mechanics | Relativity for Rose

Question

A falling stone takes 0.250 s to travel past a window 1.9 m tall. From what height above the top of the window did the stone fall?

Answer 1

Let's first find the initial velocity of the stone at the top of the window from the kinematic equation (also, let's take upwards as a positive direction):

$$x = x_0 + v_0t + \dfrac{1}{2}gt^2,$$

here, $x = 0 m$ is the final position of the stone at the bottom of the window, $x_0 = 1.9 m$ is the initial position of the stone at the top of the window, $v_0$ is the initial velocity of the stone, $g$ is the acceleration due to gravity and $t$ is the time the stone takes to travel this distance.

Then, we get:

$$v_0 = \dfrac{x - x_0 - \dfrac{1}{2}gt^2}{t},$$ $$v_0 = \dfrac{- 1.9 m - \dfrac{1}{2} \cdot (-9.8 \dfrac{m}{s^2}) \cdot (0.25 s)^2}{0.25 s} = -6.4 \dfrac{m}{s}.$$

The sign minus indicates that the velocity of the stone directed downward.

Finally, we can find from what height above the top of the wwindow the stone fall from the anothe kinematic equation:

$$v^2 = v_0^2 + 2gh,$$ $$h = \dfrac{v^2 - v_0^2}{2g} = \dfrac{(0 \dfrac{m}{s})^2 - (-6.4 \dfrac{m}{s})^2}{2 \cdot (-9.8 \dfrac{m}{s^2})} = 2.1 m.$$

Answer:

$h = 2.1 m.$