#### Answer

a. $4.0\times10^1m/s$
b. $24m/s$

#### Work Step by Step

a. Choose x = 0 and y = 0 to be the point of launch with speed v and initial angle $\theta = 0$, and upward to be the positive y direction. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward.
When the stuntperson clears the 8 cars, the position is (22 m, -1.5 m). Find the time to clear the cars from the horizontal motion at constant velocity.
$$t = \frac{22 m}{(v cos 0)}$$
Plug this expression for the time into equation 2.11b and analyze the vertical motion.
y-component of initial velocity:
$$(v)(sin 0 ^{\circ}) = 0 m/s$$
We know the final and initial heights, the initial vertical velocity, and the acceleration. Solve for the initial v = 40 m/s or $4.0\times10^1m/s$.
b. Start this problem the same way as part (a).
$$t = \frac{22 m}{(v cos 7.0 ^{\circ})}$$
Now the y-component of initial velocity is:
$$(v)(sin 7 ^{\circ}) $$
The vertical displacement is the same, and equation 2.11b is once again used. This yields a starting speed of $v = 24 m/s$.