Answer to Question #221794 in Physics for Zoe

Question

A stone is vertically thrown upward with a speed of 13 m/s from the edge of a cliff 60m high.how much later does it reach the bottom of the cliff

Answer 1

The kinematic equation of motion gives the following (y-axis directed upward, origin is at ground):

$$y = y_0 + v_0t - \dfrac{gt^2}{2}$$

where $y_0 = 60m$ is the inital coordinate (cliff height), $v_0 = 13m/s$ is the inital speed, $g = 9.81m/s^2$ is the gravitational acceleration, and $y$ is the coordinate at time $t$.

At the bottom of the cliff $y = 0$, this, substituting it, obtain:

$$0 = 60 + 13t - 4.905t^2\\ 4.905t^2 - 13t - 60 = 0$$

Solving the quadratic equation and taking the positive solution, obtain:

$$t \approx 7.7s$$

Answer. 7.7 s.