Physics Practice Problems Projectile Motion: Horizontal & Negative Launch Practice Problems Solution: You are watching an archery tournament when you st...

# Solution: 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?

###### Problem

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?

###### Solution

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:

1. Identify the target variable  and known variables for each direction—remember that only 3 of the 5 variablesx or Δy, v0, vf, a, and t) are needed for each direction. Also, it always helps to sketch out the problem and label all your known information!
2. Choose a UAM equation—sometimes you'll be able to go directly for the target variable, sometimes another step will be needed in between.
3. Solve the equation for the target (or intermediate) variable, then substitute known values and calculate the answer.

The four UAM (kinematics) equations are:

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 ay = −g, and ax = 0 (because the only acceleration acting on a projectile once it's launched is gravity.)

For a horizontally launched projectile, we also know that v0y = 0.

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

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

$\overline{)\mathbf{|}\stackrel{\mathbf{⇀}}{\mathbf{v}}\mathbf{|}{\mathbf{=}}\sqrt{{{\mathbf{v}}_{\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{v}}_{\mathbf{y}}}^{\mathbf{2}}}}$  and

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. v0x = ?
ii. Δx = 61.0 m
iii. ax = 0 m/s2
iv. vfx = ?
v. t = ?

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