Question

While performing a stunt, Danger Dog rides his motorcycle off the edge of a cliff with a velocity of 32 m/s. If the cliff is 45 m high, how far from the base of the cliff will Danger Dog land? Enter your answer in meters. 11 m 96 m 288 m 4 m

Answer 1

Answer:

The correct answer is b: 96 m.

Explanation:

The total distance in the horizontal direction traveled by Danger Dog can be calculated using the following equation:

$x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}$   (1)

Where:

$x_{f}$: is the final position in the horizontal direction =?                                      

$x_{0}$: is the initial position in the horizontal direction = 0                                      

$v_{0_{x}}$: is the initial velocity in the horizontal direction = 32 m/s

t: is the time

a: is the acceleration = 0 (he is accelerated only by gravity)

So, we need to find the time. We can find it as follows:

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

Where:

$y_{f}$: is the final position in the vertical direction = 0                                      

$y_{0}$: is the initial position in the vertical direction = 45 m                                      

$v_{0_{y}}$: is the initial velocity in the vertical direction = 0

g: is the acceleration due to gravity = 10 m/s²      

By solving equation (2) for "t" we have:

$t = \sqrt{\frac{2(y_{0} - y_{f})}{g}} = \sqrt{\frac{2*45 m}{10 m/s^{2}}} = 3 s$

Hence, the total distance traveled by Danger Dog in the x-direction is (equation 1):

$x_{f} = 32 m/s*3 s = 96 m$    

Therefore, the correct answer is b: 96 m.

I hope it helps you!