Can You Calculate the Time Needed for a Model Rocket to Pass Through a Hoop?

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a model rocket that needs to pass through a hoop located 20 meters above the launch point. The scenario includes the rocket's initial conditions, such as its mass, initial horizontal speed, and the thrust produced by its engine.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the rocket's vertical and horizontal motion, questioning the assumptions made about acceleration and the equations used to calculate time and distance. There is discussion about the correct interpretation of thrust and its effect on acceleration.

Discussion Status

The conversation is ongoing, with participants attempting to clarify the problem's requirements and the physics involved. Some have provided calculations and reasoning, while others are questioning the validity of the approaches taken and the assumptions made regarding forces and motion.

Contextual Notes

There is some confusion regarding the application of thrust as it relates to acceleration and the equations of motion. Participants are also navigating the distinction between vertical and horizontal components of the rocket's trajectory.

quick
Messages
42
Reaction score
0
A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exerts an 8.0 N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point.

i already have the answer of 7.62 but i need to know how to do it. could someone explain how to get the solution?
 
Physics news on Phys.org
i realize that the acceleration for the x component is 0. the acceleration in y component is just 8 / .5 = 16 m/s^2
the position at time t is
x = v_0*t = 3.0t
y = .5* A_y*t^2 = 8*t^2
im trying to get it to pass a hoop that is 20 m high so i tried 20 = 8*t^2 and i got a time of 1.58 s. i tried putting that time back into the x position equation but i only get 4.74. what am i doing wrong?
 
What is the question asking for? You just sort of gave us a situation and a goal but not what unknown conditions we need to solve for. Are you asking for at what distance away from the hoop horizontally the thrusters should be turned on at?

Why would 20 = 8*t^2? if 8 is thrust in N then your units are like not position but rather mass * position. The vertical component of displacement is given by d = 1/2*a*t^2. If 8 is the force what is the acceleration?
 
Last edited by a moderator:
thats exactly what i want to find out. but i already know the distance, i need to find how to get that answer.
 
(read above post)
 
ok i got acceleration by force/mass = 8 / .5 = 16 m/s^2
and then using the equation y = y_0 + v_0(t - t_0) + 1/2(a(t-t_0)^2
so i set 20 as y since that is the distance the hoop is above the ground
so 20 = 0 + 0 + 1/2 (16) t^2
so t = sqrt(20/8)
 
however by using the answer and the x position formula,
7.62 = 3t
t = 2.54
but obviously i won't be able to get that using the answer
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 2 ·
Replies
2
Views
8K
Replies
7
Views
10K
Replies
9
Views
26K
  • · Replies 2 ·
Replies
2
Views
19K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
1
Views
3K