2 The mountain lion can jump to a height of $h = 12.0\U{ft}$ when
3 leaving the ground at an angle of $\theta = 45.0\dg$. With what
4 speed, in SI units, does it leave the ground to make the leap?
5 \end{problem*} % problem 3.9
8 First, we'll convert the height into SI units:
10 h = 12.0\U{ft} \p[{ \frac{1\U{m}}{3.28\U{ft}} }]
14 Next, arrange the information we know in a table, calling the launch
15 point $P_0$ and the peak point $P_1$.\\
16 \begin{tabular}{r||r|r|}
17 Point & $P_0$ & $P_1$ \\
20 $a_x$ & \multicolumn{2}{|c|}{$0\U{m}$} \\
22 $a_y$ & \multicolumn{2}{|c|}{$-9.8\U{m}$} \\
24 $v_x$ & \multicolumn{2}{|c|}{?} \\
26 $v_y$ & ? & $0\U{m/s}$ \\
30 $y$ & $0\U{m}$ & $3.66\U{m}$ \\
35 Where we know $v_1 = 0\U{m/s}$ because $P_1$ is at the apex of the jump
36 and $x_0$, $y_0$, and $t_0$ through our choice of coordinate frame.
38 We want to pick an equation to tell us something about the initial velocity
39 (because they told us $\theta$, either $v_{x0}$ or $v_{y0}$ will suffice.).
40 Looking at our 4 constant acceleration equations (text p. 53),
42 v_{xf} &= v_{xi} + a_x t \\
43 x_f &= x_i + \frac{1}{2}(v_{xf} + v_{xi}) t \\
44 x_f &= x_i + v_{xi} t + \frac{1}{2} a_x t^2 \label{eqn.t_sqr}\\
45 v_{xf}^2 &= v_{xi}^2 + 2 a_x (x_f - x_i) \label{eqn.v_sqr}
47 We see that eqn.~\ref{eqn.v_sqr} applied to the $y$ direction is
48 perfect, because it has no information in it that we don't already
51 v_{y1}^2 &= (0\U{m/s})^2 = v_{y0}^2 + 2 a_y (y_1 - y_0) \\
52 v_{y0}^2 &= -2 a_y y_1 \\
53 v_{y0} &= \sqrt{ -2 a_y y_1 }
55 And we use the angle to solve for the magnitude of the inital velocity:
57 v_{y0} &= v_0 \sin \theta \\
58 v_0 &= \frac{v_{y0}}{\sin \theta}
59 = \frac{\sqrt{ -2 a_y y_1 }}{\sin \theta}
60 = \frac{\sqrt{-2 \cdot (-9.8\U{m/s}^2) \cdot 3.66\U{m}}}{\sin 45^o}