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## Homework Statement

A spring is fired from a launcher angled Θ degrees from the floor. The spring has force constant k and mass M. The target is Δd metres away from the launcher and the target is elevated Δy metres. When the spring is launched, it travels at initial speed v0 towards the target, Θ degrees from the horizontal. The acceleration due to gravity is g = -9.8 m/s

^{2}. What is the distance the spring needs to be pulled back, x?

## Homework Equations

t = time

x = deformation of spring

Ee = 0.5kx

^{2}

Ek = 0.5mv0

^{2}

Δy = (t)(v0)(sinΘ) + 0.5(g)(t

^{2})

Δd = (v0)(cosΘ)(t)

## The Attempt at a Solution

I just need someone to check my calculations. I need to figure out how far to pull back the spring based on a given angle, and a given horizontal distance and vertical distance to the target. There are no specific numbers, I just need a general equation.

The elastic potential energy in the spring is approximately equal to the kinetic energy the spring has after the launch. Air resistance and friction are negligible.

Therefore, 0.5kx

^{2}=0.5mv0

^{2}

kx

^{2}= mv0

^{2}

v0

^{2}= (kx

^{2}) / m

Rearranging Δd = (v0)(cosΘ)(t)

t = (Δd) / (v0 cosΘ)

Substitute t into Δy = (t)(v0)(sinΘ) + 0.5(g)(t

^{2})

Δy = (Δdv0sinΘ/v0cosΘ)(v0)(sinΘ) + 0.5(g)(Δd/v0cosΘ)

^{2}

Δy = ΔdtanΘ + (gΔd

^{2})/(2v0

^{2}cos

^{2}Θ)

Δy - ΔdtanΘ = (gΔd

^{2})/(2v0

^{2}cos

^{2}Θ)

2v0

^{2}cos

^{2}Θ = (gΔd

^{2})/(Δy - ΔdtanΘ)

v0

^{2}= (gΔd

^{2})/(2Δycos

^{2}Θ-2ΔdsinΘcosΘ)

v0

^{2}= (gΔd

^{2})/(2Δycos

^{2}Θ-Δdsin2Θ)

Substitute v0

^{2}= (kx

^{2}) / m

(kx

^{2}) / m = (gΔd

^{2})/(2Δycos

^{2}Θ-Δdsin2Θ)

x

^{2}= (gΔd

^{2}m)/(2kΔycos

^{2}Θ-Δdksin2Θ)

x = √(gΔd

^{2}m)/(2kΔycos

^{2}Θ-Δdksin2Θ)

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