Answer to Question #143150 in Physics for DeAndre

Question

A car drives straight off the edge of a cliff that is 54 m high. The police at the scene of the accident note
that the point of impact is 130 m from the base of the cliff. How fast was the car traveling when it went
over the cliff?

Answer 1

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

$$x=v_0t, (1)$$ $$y=\dfrac{1}{2}gt^2, (2)$$

here, $x=130\ m$ is the horizontal displacement of the car (or the distance from the base of the cliff), $v_0$ is the initial velocity of the car when it went over the cliff, $t$ is the time that the car takes to reach the ground, $y=54\ m$ is the vertical displacement of the car (or the height of the cliff), $g = 9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

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

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

Then, we can substitute $t$ into the first equation and find $v_0$:

$$x=v_0\sqrt{\dfrac{2y}{g}},$$ $$v_0=\dfrac{x}{\sqrt{\dfrac{2y}{g}}}=\dfrac{130\ m}{\sqrt{\dfrac{2\cdot 54\ m}{9.8\ \dfrac{m}{s^2}}}}=39\ \dfrac{m}{s}.$$

Answer:

$v_0=39\ \dfrac{m}{s}.$