Equation relating launch velocity to distance pulled back.

In summary, the conversation is about designing a projectile launcher using elastic bands or springs and determining the equation that links launch velocity to the distance the elastic band/spring is pulled back. The attempted solution involves measuring the force applied to pull back the projectile, using Hooke's Law and the kinetic energy equation to determine the launch velocity, and finding the energy stored in the launcher to be proportional to the distance pulled back. Further clarification and assistance is needed in this process.
  • #1
sharky939
2
0

Homework Statement



I've been tasked with designing a projectile launcher that must be built using either elastic bands or springs. The objective is to measure the launch velocity against the distance the projectile is pulled back before release. The launcher itself isn't supposed to be very large, as the projectile is quite small.
Basically, I need help in designing a very simple launcher, but my main need of help is discovering the equation that links launch velocity to the distance the elastic band/spring is pulled back.

Homework Equations





The Attempt at a Solution



My best attempt so far at the equation starts with measuring the force I aplly to pull back the projectile a certain distance

Then work out acceleration = force/mass (But I don't know what mass to use!)
Which then leads to final velocity (0) = Launch velocity (squared) + 2*acceleration*distance pulled back.

Any help would be greatly appreciated
 
Physics news on Phys.org
  • #2
Right, I managed to make Force equal to Hooke's Law and the kinetic energy equation, KE = 1/2mv^2.
If I get the answer in Joules from this equation, which is equal to Newton metres, and I divide through by the distance pulled back, surely then the I'd get a value in Newtons, which would be a force.
That would then mean

F = kx = 1/2MV^2
Are there any holes in this method I've fallen into?? My value of V is the lauch velocity, worked out by using the horizontal component of the projectile motion of the projectile.
 
  • #3
Energy is force * distance so the energy stored in the lanucher is average force (hookes law) * distance pulled back, this will equal the KE of the object at launch, which to a first approximation is proprtional to the distance.
 

Related to Equation relating launch velocity to distance pulled back.

1. How does the launch velocity affect the distance pulled back?

The launch velocity and distance pulled back have a direct relationship, meaning that as the launch velocity increases, the distance pulled back also increases. This is because the launch velocity is the initial speed at which an object is launched and determines how far the object will travel.

2. What is the equation to calculate the launch velocity to distance pulled back?

The equation is: d = (v2sin2θ)/g, where d is the distance pulled back, v is the launch velocity, θ is the launch angle, and g is the acceleration due to gravity.

3. How does the launch angle affect the distance pulled back?

The launch angle also has a direct effect on the distance pulled back. As the launch angle increases, the distance pulled back also increases. This is because the launch angle determines the direction of the launch, and a higher angle means the object will travel a longer horizontal distance before reaching the ground.

4. Can the distance pulled back be greater than the launch velocity?

No, the distance pulled back cannot be greater than the launch velocity. This is because the launch velocity is the maximum speed at which the object is launched and determines the maximum distance it can travel.

5. How can the launch velocity to distance pulled back equation be applied in real life?

This equation can be applied in various real-life scenarios where an object is launched, such as in sports like javelin throwing or long jump. It can also be used in projectile motion calculations in physics or engineering. By knowing the launch velocity and angle, we can predict the distance the object will travel and make adjustments to improve its performance.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
514
  • Introductory Physics Homework Help
Replies
21
Views
289
  • Introductory Physics Homework Help
Replies
23
Views
347
  • Introductory Physics Homework Help
Replies
5
Views
1K
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
21K
  • Introductory Physics Homework Help
Replies
4
Views
518
  • Introductory Physics Homework Help
Replies
15
Views
8K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
Back
Top