# A car drives straight off the edge of a cliff that is 58 m high. The police at the scene of the accident note that the point of impact is 0.136 km from the base of the cliff. How fast was the car traveling when it went over the cliff?

Sep 17, 2015

$40 \text{m/s}$

#### Explanation:

To get the time of flight we can use the fact that as the car falls over the cliff the initial vertical component of its velocity is zero.

It then falls under gravity with an acceleration of $g$.

So:

$s = \frac{1}{2} \text{g} {t}^{2}$

${t}^{2} = \frac{2 s}{g}$

$t = \sqrt{\frac{2 s}{g}}$

$t = \sqrt{\frac{2 \times 58}{9.8}}$

$t = 3.44 \text{s}$

Now we know the time of flight, we can get the horizontal component of velocity as it went over the cliff since this is constant:

$v = \frac{\Delta s}{\Delta t} = \frac{136}{3.44} = 40 \text{m/s}$

I think the police should fence off this cliff as there seems to be a whole lot of vehicles driving off it.