Question

A Chevy Camaro drives straight off the top level off a parking garage at 9 m/s. If the car landed 81 meters away from the base of the parking garage, how high (height) was the top level? 196.9 meters 296.9 meters 396.9 meters 496.9 meters​

Answer 1

Answer:

The correct option is c: 396.9 meters.

Explanation:  

First, we need to find the time at which the car landed:

$v = \frac{x}{t}$

$t = \frac{x}{v} = \frac{81 m}{9 m/s} = 9 s$                                                                    

Now, we can find the height of the top-level:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Since the car has only a velocity in the horizontal direction, we have:

$0 = y_{0} + 0*t - \frac{1}{2}gt^{2}$

$y_{f} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(9 s)^{2} = 397 m$

Therefore, the correct option is c: 396.9 meters.

I hope it helps you!