Answer to Question #152095 in Physics for Emma

Question

A stone is thrown horizontally at a speed of 86.3m/s from the top of a cliff 10.6m high.

What is the horizontal distance from the base of the cliff where the stone lands.

*hint* you will have to solve for time in the cortical direction friest before you can solve for the answer.

Answer 1

Time of falling $t$ can be found from the kinematic equation (if there is no vertical initial velocity component):

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

where $h = 10.6m$ is the height, and $g = 9.81m/s^2$ is the gravitational acceleration. Thus, obtain:

$$t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2\cdot 10.6}{9.81}} \approx 1.47s$$

In this time the stone covers the following horizontal distance:

$$d = v_0t = 86.3m/s\cdot 1.47s \approx 126.9m$$

Here $v_0 = 86.3m/s$ is the initial horizontal velocity of the stone.

Answer. 126.9 m.