Question

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 36.0-m-tall building. The rocket's engine produces a horizontal acceleration of (1.60 m/s3)t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g, downward. Air resistance can be neglected. What horizontal distance does the rocket travel before reaching the ground

Answer 1

Answer:

48.4 m

Explanation:

To calculate the horizontal distance we need to find the flight time:

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

Where:

$y_{f}$ is the final height

$y_{0}$ is the initial height

$v_{0y}$ is the initial vertical speed

t is the time

g is the gravity

$0 = 36.0 m - \frac{1}{2}9.81t^{2}$

$t = \sqrt{\frac{2*36.0 m}{9.81 m/s^{2}}} = 2.71 s$

Now, we can find the distance:

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

$x_{f} = 12.0 m/s*2.71 s + \frac{1}{2}1.60 m/s^{3}(2.71 s)^{3} = 48.4 m$

Therefore, the horizontal distance traveled by the rocket is 48.4 m.

I hope it helps you!