Spring launcher equation for x so it hits the target

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To determine how far to pull back the spring for it to hit the target, the discussion emphasizes using conservation of energy and projectile motion equations. The mass of the spring is 0.024 kg, with a spring constant of 25 N/m, and the launch angle is set at 40 degrees. Trials indicate varying distances based on how far the spring is pulled back, with recorded distances ranging from 0.52 m to 5.37 m. The user struggles with calculating the velocity needed for the equations, highlighting the need for a relationship between initial speed, acceleration, time, and height difference. The target is confirmed to be at floor level, which is crucial for the calculations.
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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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Is the target at floor level?
You need an equation relating to vertical motion, relating initial speed, acceleration, time and height difference.
 
yes it's on the floor
 

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