Problem: A dart is thrown horizontally with an initial speed of 10 m/s toward point  P, the bull’s-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later.(a)What is the distance PQ? (b) How far away from the dart board is the dart released?

This problem asks us to calculate the vertical and horizontal distance the dart travels, given the initial velocity and time of flight.

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 variables (Δx or Δy, v₀, 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!
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:

$$\begin{gathered} \boxed{\mathbf{ }{\mathbf{v}}_{\mathbf{f}}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{v}}_{\mathbf{0}}\mathbf{ }\mathbf{+}\mathbf{a}\mathbf{t} \\ \mathbf{∆}\mathbf{x}\mathbf{=}\mathbf{ }\mathbf{\left(}\frac{{\mathbf{v}}_{\mathbf{f}}\mathbf{+}{\mathbf{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}\mathbf{t} \\ \mathbf{∆}\mathbf{x}\mathbf{=}\mathbf{ }{\mathbf{v}}_{\mathbf{0}}\mathbf{t}\mathbf{+}\frac{1}{2}\mathbf{a}{\mathbf{t}}^{\mathbf{2}} \\ \mathbf{ }{{\mathbf{v}}_{\mathbf{f}}}^{\mathbf{2}}\mathbf{=}\mathbf{ }{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{∆}\mathbf{x}} \end{gathered}$$

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 after a projectile is launched, the only acceleration is due to gravity.)

For a horizontally launched projectile, we also know that v_0y = 0.