You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 61.0 m away, making a 3.00° angle with the ground.

How fast was the arrow shot?

In this problem, we're asked to calculate the **initial speed**, given the **horizontal distance travelled** and the initial and final **directions** of motion.

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

- Identify the
and__target variable__for each direction—remember that__known variables__*only*(Δ**3**of the**5**variables*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__—sometimes you'll be able to go directly for the target variable, sometimes another step will be needed in between.**equation**for the target (or intermediate) variable, then**Solve**the equation__substitute known values__and__calculate__the answer.

The four UAM (kinematics) equations are:

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

We define our coordinate system so that the **+ y-axis is pointing upwards** and the

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:

$\overline{)\begin{array}{rcl}{\mathbf{v}}_{\mathbf{x}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\theta}\\ {\mathbf{v}}_{\mathbf{y}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\theta}\end{array}}$

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

$\overline{)\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}{\mathbf{=}}\sqrt{{{\mathbf{v}}_{\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{v}}_{\mathbf{y}}}^{\mathbf{2}}}}$ and $\overline{){\mathbf{t}}{\mathbf{a}}{\mathbf{n}}{\mathbf{}}{\mathbf{\theta}}{\mathbf{=}}\frac{{\mathbf{v}}_{\mathbf{y}}}{{\mathbf{v}}_{\mathbf{x}}}}$

**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. **v**_{0x} = ?

ii. **Δ****x**** = 61.0 m**

iii. **a _{x}**

iv.

v.

Projectile Motion: Horizontal & Negative Launch

Projectile Motion: Horizontal & Negative Launch

Projectile Motion: Horizontal & Negative Launch

Projectile Motion: Horizontal & Negative Launch

Projectile Motion: Horizontal & Negative Launch