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`

Step-by-step explanation:

For this situation we will use the following equations:

`y=y_(o)+V_(o)t+(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=(V_(1)+V_(2))/(2)` (3)

For this rocket is:

`V=(27 m/s)/(2)=13.5 m/s` (4)

Time is determined by:

`t=(y)/(V)` (5)

`t=(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=(1)/(2)at^(2)` (8)

We will use
`y=4.2 m` :

`4.2 m=(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=(1)/(2)at^(2)` but for
`t=0.2 s`:

`y=(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)