In this problem, we're asked to calculate the **initial speed** an arrow is released with, given the **horizontal distance traveled** and the initial and final **directions** of motion.

For **projectile motion problems in general**, we'll follow these steps to solve:

- Identify the
__target variable__ and __known variables__ for each direction—remember that *only ***3** of the **5** variables (Δ*x* or Δ*y*, *v*_{0}, *v*_{f}, *a*, and *t*)* are needed* for each direction. Also, it always helps to sketch out the problem and label all your known information! __Choose a UAM __**equation**—sometimes you'll be able to go directly for the target variable, sometimes another step will be needed in between.**Solve** the equation for the target (or intermediate) variable, then __substitute known values__ and __calculate__ the answer.

The four UAM (kinematics) equations are:

$\overline{){{v}}_{{f}}{}{=}{}{{v}}_{{0}}{}{+}{a}{t}\phantom{\rule{0ex}{0ex}}{\u2206}{x}{=}{}\left(\frac{{v}_{f}+{v}_{0}}{2}\right){t}\phantom{\rule{0ex}{0ex}}{\u2206}{x}{=}{}{{v}}_{{0}}{t}{+}{\frac{1}{2}}{a}{{t}}^{{2}}\phantom{\rule{0ex}{0ex}}{}{{{v}}_{{f}}}^{{2}}{=}{}{{{v}}_{{0}}}^{{2}}{}{+}{2}{a}{\u2206}{x}}$

We define our coordinate system so that the **+***y*-axis is pointing upwards and the **+***x*-direction is **horizontal along the launch direction**. That means **a**_{y}** = −****g**, and **a**_{x}** = 0** (because the only acceleration acting on a projectile once it's launched is gravity.)

For a **horizontally launched projectile**, we __also__ know that *v*_{0}_{y} = 0.

Finally, we might need to know how to decompose a velocity vector into its *x*- and *y*-components:

$\overline{)\begin{array}{rcl}{v}_{x}& {=}& \left|\stackrel{\rightharpoonup}{v}\right|\mathrm{cos}\theta \\ {v}_{y}& {=}& \left|\stackrel{\rightharpoonup}{v}\right|\mathrm{sin}\theta \end{array}}$

Or how to get the magnitude and angle of a velocity vector if we know the components:

$\overline{)\left|\stackrel{\rightharpoonup}{v}\right|{=}\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}}$ and $\overline{){\mathrm{tan}}{}{\theta}{=}\frac{{A}_{y}}{{A}_{x}}}$

**Step 1****.** The problem gives us the __horizontal distance the arrow traveled__ and the __angle of its final velocity__, as well as the fact that it's shot horizontally. So our known, unknown, and target variables are as follows:

In the *x* direction:

i. **Δ****x**** = 61.0 m** (known)

ii. **v**_{0x} = ? (target)

iii. *v*_{fx} = ? (unknown)

iv. *t* = ? (unknown)

v. **a**_{x}** = 0 m/s**^{2} (known)