Answer:
48.4 m
Explanation:
To calculate the horizontal distance we need to find the flight time:
[tex] y_{f} = y_{0} + v_{0y}*t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex]y_{f}[/tex] is the final height
[tex]y_{0}[/tex] is the initial height
[tex]v_{0y}[/tex] is the initial vertical speed
t is the time
g is the gravity
[tex] 0 = 36.0 m - \frac{1}{2}9.81t^{2} [/tex]
[tex]t = \sqrt{\frac{2*36.0 m}{9.81 m/s^{2}}} = 2.71 s[/tex]
Now, we can find the distance:
[tex] x_{f} = x_{0} + v_{0x}*t + \frac{1}{2}at^{2} [/tex]
[tex]x_{f} = 12.0 m/s*2.71 s + \frac{1}{2}1.60 m/s^{3}(2.71 s)^{3} = 48.4 m[/tex]
Therefore, the horizontal distance traveled by the rocket is 48.4 m.
I hope it helps you!
