Q51P - University Physics with Modern Physics

Question

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0m tall building. The rocket’s engine produces a horizontal acceleration of $\left(160m/s^3\right)$t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g, downward. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?

Step-by-Step Solution

Answer

The horizontal distance the rocket travel before reaching ground 5087.53m.

Step 1: Newton’s law

The velocity contains two components, one is horizontal component and another one is vertical. 

 

According to the newton’s laws of motion,

 

$s=ut+\frac{1}{2}a{t}^2$ 

 

Where, s, u, t, a, are displacement, initial velocity, time and acceleration.

 

From the newton’s law of motion

 

v=u+at

 

Here v, u, a, t are final velocity, initial velocity, acceleration due to gravity, and time.

Step 2: Given

Initial velocity = 12.0m/s  

 

Building = 30m

 

Horizontal Acceleration = $\left(160m/s^3\right)t$  

Step 3: Horizontal distance travel by rocket before reaching the ground

For the vertical motion from the law of motion of Newton, 

$$\begin{gathered} \mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ 30\mathrm{m}=12\mathrm{m}/\mathrm{s}\times \mathrm{t}+\frac{1}{2}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times {\mathrm{t}}^2 \\ \mathrm{t}=3.979\mathrm{s} \end{gathered}$$                                                                                           

 

 

For the horizontal motion,

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ s=12m/s\times 3.979s+\frac{1}{2}\times 160m/s^3\times {\left(3.979s\right)}^3 \\ s=5087.53m \end{gathered}$$