Question

A car traveling 24.5 m/s runs over a cliff and lands 8 m away from the base. how high is the cliff?

Answer 1

Answer: 0.522 m

Explanation:

This situation is related to projectile motion, where the initial velocity of the car $V_{o}=24.5 m/s$ has only the horizontal component. This means the angle is equal to zero ($\theta=0$). And the equations we will use to find the height $y_{o}$ of the cliff are:

$y=y_{o}+V_{o}sen \theta t+\frac{1}{2}gt^{2}$ (1)  

Where:  

$y=0 m$ is the final height of the car

$y_{o}$ is the initial height

$V_{o}=24.5 m/s$ is the initial velocity of the car

$t$ is the time

$g=-9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=0$ is the angle

$x=8 m$ is the horizontal distance traveled by the car after passing the cliff

Well, firstly we have to find $t$  from (1):

$0=y_{o}+\frac{1}{2}gt^{2}$ (3)  

$t=\sqrt{-\frac{2y_{o}}{g}}$ (4)  

Substituting (4) in (2):

$x=V_{o}cos \theta (\sqrt{-\frac{2y_{o}}{g}})$ (5)  

Isolating $y_{o}$:

$y_{o}=-\frac{x^{2} g}{2V_{o}^{2}}$ (6)  

$y_{o}=-\frac{(8m)^{2} (-9.8 m/s^{2})}{2(24.5)^{2}}$ (7)  

Finally:

$y_{o}=0.522 m$