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:
[tex] x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2} [/tex] (1)
Where:
[tex] x_{f}[/tex]: is the final position in the horizontal direction =?
[tex]x_{0}[/tex]: is the initial position in the horizontal direction = 0
[tex] v_{0_{x}} [/tex]: 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:
[tex] y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} [/tex] (2)
Where:
[tex] y_{f}[/tex]: is the final position in the vertical direction = 0
[tex]y_{0}[/tex]: is the initial position in the vertical direction = 45 m
[tex] v_{0_{y}} [/tex]: 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:
[tex] t = \sqrt{\frac{2(y_{0} - y_{f})}{g}} = \sqrt{\frac{2*45 m}{10 m/s^{2}}} = 3 s [/tex]
Hence, the total distance traveled by Danger Dog in the x-direction is (equation 1):
[tex] x_{f} = 32 m/s*3 s = 96 m [/tex]
Therefore, the correct answer is b: 96 m.
I hope it helps you!
