Answer to Question #210904 in Physics for Jax

Question

A stone was dropped on top of the table, 2 meters above the ground. Find the distance of the ground level at time t? How long will it take the stone to reach the ground?

Answer 1

Let's denote the distance from the table to the ball at time $t$ as $s$, and from the ball to the ground as $h$. It is clear that $s + h = H$ at any moment, where $H = 2m$ is the height of the table.

According to the kinematic equation, have:

$$s = \dfrac{gt^2}{2}$$

where $g = 9.8N/kg$ is the gravitational acceleration. Thus the distance of the ground level is:

$$h = H-s = H - \dfrac{gt^2}{2}$$

If stone reaches the ground, $h = 0$. Substituting this into the last equation, and expressing the time, obtain:

$$0 = H-\dfrac{gt^2}{2}\\ \dfrac{gt^2}{2} = H\\ t = \sqrt{\dfrac{2H}{g}}\\ t = \sqrt{\dfrac{2\cdot 2}{9.8}} \approx 0.64s$$

Answer. $h = H - \dfrac{gt^2}{2}$, $t = 0.64s$.