Spring War

Applying the law of conservation of mechanical energy yields: $E_{k0} + U_{g0} + U_{sp0} = E_{k1} + U_{g1} + U_{sp1}$

$\frac{1}{2}kx^{2}_{C} = \frac{1}{2}mv^{2}_{0}$

Where $k$ is the spring constant, $x_{C}$ is the compression length of the spring, $m$ is the mass of the dart, and $v_{0}$ is the initial velocity of the dart when it leaves the chamber of the gun.

Solving for $v_{0}$ yields $v_{0} = x_{C}\sqrt{\frac{k}{m}}$ . If we solve for $k$ in the same equation, we get $k = \frac{mv^{2}_{0}}{x^{2}_{C}}$ . Knowing that when $x_{C} = l, v_{0} = v_{max}$ , we can solve for $k$ by substituting these values, yielding $k = \frac{mv^{2}_{max}}{l^{2}} = \frac{(0.0013 kg) * (409.44 in/s)^{2}}{(6 in)^{2}} = 6.0537 kg/s^{2}$ .

Let the trajectory of the dart be described by the coordinates $(x(t), y(t)) = (v_{0}\cos(a)t, -\frac{g}{2}t^{2} + v_{0}\sin(a)t + h_{0})$ where $a$ is the angle of elevation of the gun, $g$ is the acceleration due to gravity, and $h_{0}$ is the initial height of the dart.

The dart impacts Timmy at a time $T$ such that $x(T) = x_{1}$ where $x_{1}$ is the distance between Timmy and Jimmy (Jimmy is assumed to be at (0, $h_{0}$ ) ). Solving for $T$ yields $T = \frac{x_{1}}{v_{0}\cos(a)}$ .

At this same time $T$ , $y(T) = h_{1}$ . Substituting $T$ for $t$ in $-\frac{g}{2}t^{2} + v_{0}\sin(a)t + h_{0} = h_{1}$ and solving for $v^{2}_{0}$ yields $v^{2}_{0} = \frac{gx^{2}_{1}}{2\cos^{2}(a)(x_{1}\tan(a) + h_{0} - h_{1})}$ .

Substituting $v_{0} = x_{C}\sqrt{\frac{k}{m}}$ and solving for $x_{C}$ yields $x_{C} = \sqrt{\frac{mgx^{2}_{1}}{2k\cos^{2}(a)(x_{1}\tan(a) + h_{0} - h_{1})}}$ . Upon substituting all given values for $m, g, k, x_{1}, a, h_{0},$ and $h_{1}$ returns a value of 5.451562 in for $x_{C}$ . This is the minimum value needed to hit Timmy, as in order to hit the target one could have just compressed the spring to its maximum length (but that wouldn't have been a fun problem, would it?)