Spring launcher equation for x so it hits the target

In summary, the problem at hand is to determine the distance to pull back a spring in order to hit a target on the floor. The mass of the spring is 0.024kg with a spring constant of 25N/m. The launch angle is set at 40 degrees and the length of the spring when not stretched is 0.20m. The height of the launcher is 0.54m. To account for air resistance, the distance will be given on the day of testing. Trials have been conducted with varying lengths of spring and corresponding distances. The equations needed for the solution include conservation of energy, projectile motion, and an equation for vertical motion. The target is at floor level.
  • #1
firezap
29
0

Homework Statement


I need to find out how far to pull the spring back so it will travel to the target. The launcher and the target is both set on the floor. The spring is the projectile. I will be given the distance on the day of testing. i should take into account air resistance so i will be more accurate.

Mass of spring = 0.024kg
k = 25N/m
launch angle set at 40 degrees
length of spring when not stretched is 0.20m
height of launcher is 0.54m
My trials pulling spring back and recording distance
.25m went .52m
.3m went 1.67m
.34m went 2.2m
.4m went 4.78m
.43m went 5.37m

Homework Equations


conservation of energy
Es = Ek
1/2mv^2 = 1/2kx^2
projectile motion
d = v1xt
v1x = v1cosӨ

The Attempt at a Solution


1/2mv^2 = 1/2kx^2
i get stuck on this part. plugging in numbers but i do not have the velocity. What i do next?
 
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  • #2
Is the target at floor level?
You need an equation relating to vertical motion, relating initial speed, acceleration, time and height difference.
 
  • #3
yes it's on the floor
 

FAQ: Spring launcher equation for x so it hits the target

1. What is the spring launcher equation for x?

The spring launcher equation for x is x = v0t + (1/2)at2, where x is the distance traveled by the object, v0 is the initial velocity, t is the time, and a is the acceleration.

2. How do I calculate the initial velocity for the spring launcher?

The initial velocity can be calculated by dividing the distance to be traveled by the time it takes to travel that distance. This will give you the average velocity, which can then be used as the initial velocity in the spring launcher equation.

3. How is the spring launcher equation for x used to hit a target?

The spring launcher equation for x can be used to calculate the initial velocity needed to hit a target at a specific distance. By plugging in the desired distance for x, and the known values for t and a, the equation can be rearranged to solve for v0. This initial velocity can then be used to set up the spring launcher to hit the target.

4. Is the spring launcher equation for x affected by external factors?

Yes, the spring launcher equation for x is affected by external factors such as air resistance, friction, and the weight of the object being launched. These factors can change the acceleration and initial velocity of the object, thus affecting the accuracy of hitting the target.

5. Can the spring launcher equation for x be used for objects of any weight?

Yes, the spring launcher equation for x can be used for objects of any weight as long as the external factors are taken into consideration and the equation is adjusted accordingly. However, it is important to note that heavier objects may require a stronger spring and greater initial velocity to hit the target, while lighter objects may need a weaker spring and lower initial velocity.

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