A vertical spring launcher converts stored elastic potential energy into kinetic energy and gravitational potential
energy. The mass starts from rest while the spring is compressed by a distance \(x\). When the spring reaches its
natural length, the mass leaves the spring with exit speed \(v_{\mathrm{exit}}\).
Elastic potential energy
A spring compressed by \(x\) stores elastic potential energy:
\[
U_s=\frac12kx^2
\]
Here \(k\) is the spring constant and \(x\) is the compression distance.
Upward launch
For an upward launch, the mass rises a distance \(x\) while the spring expands. Gravity removes energy during this
launch phase. Energy conservation from the compressed position to the spring exit gives:
\[
\frac12kx^2=\frac12mv_{\mathrm{exit}}^2+mgx
\]
Solving for exit speed:
\[
\begin{aligned}
\frac12mv_{\mathrm{exit}}^2 &= \frac12kx^2-mgx \\
v_{\mathrm{exit}} &= \sqrt{\frac{kx^2}{m}-2gx}
\end{aligned}
\]
The square-root expression must be nonnegative. If it is negative, the spring cannot lift the mass all the way to the
natural-length exit point.
Maximum height after an upward launch
After leaving the spring, the mass continues upward until its kinetic energy has converted into gravitational potential
energy:
\[
\frac12mv_{\mathrm{exit}}^2=mgh_{\mathrm{extra}}
\]
\[
h_{\mathrm{extra}}=\frac{v_{\mathrm{exit}}^2}{2g}
\]
The maximum height above the compressed starting position is:
\[
H_{\max}=x+h_{\mathrm{extra}}
\]
Combining the spring phase and the free-flight phase gives the direct relation:
\[
\begin{aligned}
\frac12kx^2 &= mgH_{\max} \\
H_{\max} &= \frac{kx^2}{2mg}
\end{aligned}
\]
Solving for spring constant or compression
If a target maximum height is known, the same direct energy equation can be rearranged.
Spring constant from target height.
\[
\begin{aligned}
H_{\max} &= \frac{kx^2}{2mg} \\
k &= \frac{2mgH_{\max}}{x^2}
\end{aligned}
\]
Compression from target height.
\[
\begin{aligned}
H_{\max} &= \frac{kx^2}{2mg} \\
x &= \sqrt{\frac{2mgH_{\max}}{k}}
\end{aligned}
\]
If the target is an exit speed instead, start from the exit-speed equation:
\[
v_{\mathrm{exit}}^2=\frac{kx^2}{m}-2gx
\]
Spring constant from upward exit speed.
\[
\begin{aligned}
k &= \frac{m(v_{\mathrm{exit}}^2+2gx)}{x^2}
\end{aligned}
\]
Compression from upward exit speed.
\[
\begin{aligned}
kx^2-2mgx-mv_{\mathrm{exit}}^2 &= 0 \\
x &= \frac{mg+\sqrt{m^2g^2+kmv_{\mathrm{exit}}^2}}{k}
\end{aligned}
\]
Downward launch
In a downward launch, gravity helps the spring. The mass moves downward by \(x\) while the spring expands, so gravity
adds kinetic energy:
\[
\frac12kx^2+mgx=\frac12mv_{\mathrm{exit}}^2
\]
\[
\begin{aligned}
v_{\mathrm{exit}} &= \sqrt{\frac{kx^2}{m}+2gx}
\end{aligned}
\]
If the mass then falls an additional distance \(H\) below the exit point, the impact speed is:
\[
\begin{aligned}
v_{\mathrm{impact}}^2 &= v_{\mathrm{exit}}^2+2gH \\
v_{\mathrm{impact}} &= \sqrt{v_{\mathrm{exit}}^2+2gH}
\end{aligned}
\]
Worked example
A mass \(m=0.8\ \mathrm{kg}\) is launched upward by a spring with \(k=1200\ \mathrm{N/m}\) compressed by
\(x=0.25\ \mathrm{m}\). Use \(g=9.81\ \mathrm{m/s^2}\).
Step 1: spring energy.
\[
\begin{aligned}
U_s &= \frac12kx^2 \\
&= \frac12(1200)(0.25)^2 \\
&= 37.5\ \mathrm{J}
\end{aligned}
\]
Step 2: gravitational energy needed to reach the exit.
\[
\begin{aligned}
mgx &= (0.8)(9.81)(0.25) \\
&= 1.96\ \mathrm{J}
\end{aligned}
\]
Step 3: exit kinetic energy.
\[
\begin{aligned}
K_{\mathrm{exit}} &= U_s-mgx \\
&= 37.5-1.96 \\
&= 35.5\ \mathrm{J}
\end{aligned}
\]
Step 4: exit speed.
\[
\begin{aligned}
v_{\mathrm{exit}} &= \sqrt{\frac{2K_{\mathrm{exit}}}{m}} \\
&= \sqrt{\frac{2(35.5)}{0.8}} \\
&\approx 9.43\ \mathrm{m/s}
\end{aligned}
\]
Step 5: maximum height above the compressed start.
\[
\begin{aligned}
H_{\max} &= \frac{kx^2}{2mg} \\
&= \frac{(1200)(0.25)^2}{2(0.8)(9.81)} \\
&\approx 4.78\ \mathrm{m}
\end{aligned}
\]
Formula summary
| Concept |
Formula |
Meaning |
| Spring energy |
\(U_s=\frac12kx^2\) |
Energy stored in the compressed spring. |
| Upward exit speed |
\(v_{\mathrm{exit}}=\sqrt{\frac{kx^2}{m}-2gx}\) |
Speed after gravity removes energy during the guide phase. |
| Maximum upward height |
\(H_{\max}=\frac{kx^2}{2mg}\) |
Highest point above the compressed starting position. |
| Spring constant from height |
\(k=\frac{2mgH_{\max}}{x^2}\) |
Required stiffness for a desired maximum height. |
| Compression from height |
\(x=\sqrt{\frac{2mgH_{\max}}{k}}\) |
Required compression for a desired maximum height. |
| Downward exit speed |
\(v_{\mathrm{exit}}=\sqrt{\frac{kx^2}{m}+2gx}\) |
Speed when gravity assists the launch. |
| Downward impact speed |
\(v_{\mathrm{impact}}=\sqrt{v_{\mathrm{exit}}^2+2gH}\) |
Speed after an additional free-fall drop. |
The key idea is energy conversion: spring energy becomes kinetic energy and gravitational potential energy.
