A fireworks rocket is launched vertically upward at 40 m/s

  • Thread starter BMW25
  • Start date
  • Tags
    Rocket
In summary, the first fragment reaches the ground 2.51 seconds after the explosion, while the second fragment reaches the ground t1= 2.51s after the explosion, but I don't know the time between the explosion and when the second fragment reaches the ground.
  • #1
BMW25
4
0
A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1= 2.51s after the explosion.
When does the second reach the ground?
t= ...??

logiclly, it is 2.51 s also ...but I don't know. so could you pls guys help with that out ??
 
Physics news on Phys.org
  • #2
Ok, find where it is at its peak. At its peak its velocity is zero. Use conservation of momentum to get the initial directions and speeds. That is, let the first fragment travel in some initial direction with an angle theta to the horizontal (or whatever you like), then get the other fragment initial direction and speed, and solve the equations of motion.
 
  • #3
thnx dude for your post. but could you pls explain a little bit more for me?
 
  • #4
Ok, use the equations of motion to find where (at what height) the rocket reaches its peak. Then define a set of axes: to the right horizontally is you +x axis, and, say, vertically upwards is your +y axis. Then say immediately after the explosion, the first fragment leaves at an angle theta relative to the +x axis (in the usual way we measure angles, in the anti-clockwise direction), with a speed v.
Then you can use the equations of motion to find this angle theta and the speed v, since we know it took 2.51 seconds to reach the ground.
Now we have the initial direction and speed the first fragment had. Also, initially, linear momentum was conserved, so the other fragment flew off in the opposite direction but with the same speed (because their masses are the same). You know the angle the first one flew off in, so you can find the angle the other one flew off in, because the total momentum initially must have been zero.
 

1. What is the initial velocity of the fireworks rocket?

The initial velocity of the fireworks rocket is 40 m/s, as stated in the given information.

2. Will the fireworks rocket eventually stop and fall back down?

Yes, the fireworks rocket will eventually stop and fall back down due to the force of gravity acting on it.

3. How high will the fireworks rocket reach?

To determine the height reached by the fireworks rocket, we can use the equation h = ut + (1/2)at^2, where h is the height, u is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time. Assuming no air resistance, the rocket will reach a maximum height of 80 meters after 2 seconds (when it reaches its peak) before falling back down.

4. What factors may affect the flight of the fireworks rocket?

The flight of the fireworks rocket may be affected by factors such as air resistance, wind speed and direction, and the shape and weight of the rocket.

5. Can the velocity of the fireworks rocket be changed during its flight?

Yes, the velocity of the fireworks rocket can be changed during its flight. It may be affected by external forces such as air resistance or by the rocket's own propulsion system. However, the initial velocity of 40 m/s will remain constant unless acted upon by an external force.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
895
  • Introductory Physics Homework Help
Replies
13
Views
897
  • Introductory Physics Homework Help
Replies
1
Views
695
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
22
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top