Answer to Question #140675 in Physics for joe

Question

A ball is kicked horizontally at 8.0 m/s from a cliff 80m high. How far from the base of the cliff will the stone strike the ground?

Answer 1

Let's first write the equations of motion of the ball in horizontal and vertical directions:

$$x=v_{0x}t,$$ $$y=v_{0y}t+\dfrac{1}{2}gt^2,$$

here, $x$ is the horizontal displacement of the ball (or the distance from the base of the cliff to the ball), $v_{0x}=8.0\ \dfrac{m}{s}$ is the initial horizontal velocity of the ball, $t$ is the time that the ball takes to reach the ground, $y=80\ m$ is the vertical displacement of the ball (or the height of the cliff), $v_{0y}=0\ \dfrac{m}{s}$ is the initial vertical velocity of the ball and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's find the time that the ball takes to reach the ground from the second equation:

$$t=\sqrt{\dfrac{2y}{g}}.$$

Finally, we can substitute $t$ into the first equation and find the distance from the base of the cliff to the ball:

$$x=v_{0x}t=v_{0x}\sqrt{\dfrac{2y}{g}},$$ $$x=8.0\ \dfrac{m}{s}\cdot\sqrt{\dfrac{2\cdot 80\ m}{9.8\ \dfrac{m}{s^2}}}=32\ m.$$

Answer:

$x=32\ m.$