Find Maximum Elevation of Test Rocket

  • Thread starter Thread starter jenc305
  • Start date Start date
  • Tags Tags
    Maximum
Click For Summary
The test rocket problem involves two phases: during acceleration and after the motor shuts off. Initially, the rocket reaches a height of 160 meters with a velocity of 80 m/s after 4 seconds of acceleration at 20 m/s². Once the motor stops, the rocket continues to coast upward until it reaches a maximum elevation of approximately 487 meters, calculated using projectile motion equations. Additionally, a question about horizontal projectile motion highlights confusion regarding the increasing distance between two packages dropped from a plane, which occurs due to their constant horizontal velocity while falling under gravity. Understanding these concepts is crucial for solving both vertical and horizontal motion problems effectively.
jenc305
Messages
16
Reaction score
0
This problem seems simple but I'm just not getting the correct answer, please help thank you.

A test rocket is fired straigt up from rest with a net accerleration of 20m/s^2. After 4 seconds the motor turns off, but the rocket continues to coast upward. What maximum elevation does the rocket reach?

This is what I get

X_0=0, a=20m/s^2, t=4 s

1/2(20m/s^2)(4 s)^2=160m (the correct answer is 487m :cry: )
 
Last edited:
Physics news on Phys.org
You need to think of it as 2 separate problems.

1. Rocket under the stated acceleration

2. Rocket after acceleration stops.

The final conditions of problem 1 (i.e. height and velocity) are the initial conditions for problem 2.
 
Problem 1
x=160m and v=80m/s

Problem 2
x_0=160m, v_0=80m/s, a=?
if it stopped accelerating, I would assume that the acceleration would be decreasing as the rocket coasted upward. Would this be correct?
 
So far, so good. You have the initial conditions for problem #2. Now, you need the equation of motion (in the vertical direction) for a projectile with initial conditions of distance and velocity. You must have been given such an equation or taught how to derive it using F=ma.
 
Last edited:
Thank you. This is what I got

160m-[0-(80m/s)^2/2(9.8m/s^2)]=486.5m

Also, I had another question about horizontal projectile motion.
Why does the distance of two packages steadily increase as they fall out of a plane going at constant speed.

I thought the distance would remain constant between the two packages as they fall when air friction is negligible. I can't make sense of it.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
13
Views
2K
  • · Replies 53 ·
2
Replies
53
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
40
Views
7K