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?

We can use the fact that the car's initial vertical component of velocity is zero as it falls over the cliff to determine the time of flight.

Thus:

Given that we now know the flight time, we can calculate the velocity's horizontal component as it passed over the cliff because it is constant:

Since so many cars appear to be driving off this cliff, I believe the police should fence it off.

To find the initial speed of the car when it went over the cliff, we can use the equation of motion for an object in free fall:

$s = ut + \frac{1}{2}at^2$

where:

• $s$ is the distance fallen (height of the cliff in this case),
• $u$ is the initial velocity of the car (which we want to find),
• $t$ is the time taken to fall (which we'll calculate),
• $a$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$ downward).

We can rearrange the equation to solve for $u$:

$u = \sqrt{2as}$

Given that the height of the cliff ($s$) is 58 m, and the acceleration due to gravity ($a$) is $9.8 \, \text{m/s}^2$, we can plug in these values to find $u$.

Next, we need to find the time taken to fall ($t$). We can use the horizontal distance from the base of the cliff to the point of impact (0.136 km) and the initial speed ($u$) to calculate the time taken using the formula:

$t = \frac{\text{horizontal distance}}{\text{initial speed}}$

Once we have $t$, we can plug it into the equation for $u$ to find the initial speed of the car when it went over the cliff.