Question

A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does it take for the rocket to reach this height? What was the magnitude of the rocket's acceleration? Find the height of the rocket 0.20 s after launch. Find the speed of the rocket 0.20 s after launch.

Answer 1

Answers:

a) $t=0.311 s$

b) $a=86.847 m/s^{2}$

c) $y=1.736 m$

d) $V=17.369 m/s$

Explanation:

For this situation we will use the following equations:

$y=y_{o}+V_{o}t+\frac{1}{2}at^{2}$ (1)  

$V=V_{o} + at$ (2)  

Where:  

$y$ is the height of the model rocket at a given time

$y_{o}=0$ is the initial height of the model rocket

$V_{o}=0$ is the initial velocity of the model rocket since it started from rest

$V$ is the velocity of the rocket at a given height and time

$t$ is the time it takes to the model rocket to reach a certain height

$a$ is the constant acceleration due gravity and the rocket's thrust

a) Time it takes for the rocket to reach the height=4.2 m

The average velocity of a body moving at a constant acceleration is:

$V=\frac{V_{1}+V_{2}}{2}$ (3)

For this rocket is:

$V=\frac{27 m/s}{2}=13.5 m/s$ (4)

Time is determined by:

$t=\frac{y}{V}$ (5)

$t=\frac{4.2 m}{13.5 m/s}$ (6)

Hence:

$t=0.311 s$ (7)

b) Magnitude of the rocket's acceleration

Using equation (1), with initial height and velocity equal to zero:

$y=\frac{1}{2}at^{2}$ (8)  

We will use $y=4.2 m$ :

$4.2 m=\frac{1}{2}a(0.311)^{2}$ (9)  

Finding $a$:

$a=86.847 m/s^{2}$ (10)  

c) Height of the rocket 0.20 s after launch

Using again $y=\frac{1}{2}at^{2}$ but for $t=0.2 s$:

$y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}$ (11)

$y=1.736 m$ (12)

d) Speed of the rocket 0.20 s after launch

We will use equation (2) remembering the rocket startted from rest:

$V= at$ (13)  

$V= (86.847 m/s^{2})(0.2 s)$ (14)  

Finally:

$V=17.369 m/s$ (15)