Lesson 4 · 40 min
Projectile Motion
A kicked ball, a jet of water from a fire hose and a package dropped from a drone all follow the same rule: steady motion sideways, constant acceleration downward. Rectangular components turn every projectile problem into two simple straight-line problems.
Learning objectives
- State the assumptions of the projectile model, and write \(a_x = 0\), \(a_y = -g\).
- Write and use the equations for \(x(t)\), \(y(t)\), \(v_x\) and \(v_y\) for a launch at speed \(v_0\) and angle \(\theta_0\).
- Find the time of flight, maximum height and range, including launches from a height.
- Use the trajectory equation to find the launch angle needed to hit a target.
The projectile model
A projectile is a particle that, once launched, moves under gravity alone. The model makes three assumptions:
- No air resistance. Good for dense, slow objects over short distances (a shot put, a stone, a water jet close to the nozzle); poor for a golf ball or a badminton shuttle.
- Constant gravity \(g = 9.81\ \text{m/s}^2\), straight down: the flight is short and low compared with the size of the Earth.
- A particle: the object's size and spin do not matter.
Take \(x\) horizontal and \(y\) vertically up. Then the acceleration is the same at every instant of the flight:
Projectile acceleration
\[ a_x = 0, \qquad a_y = -g \]The horizontal motion has constant velocity; the vertical motion has constant acceleration. They are independent: they share only the time \(t\).
Launch from \((x_0, y_0)\) with speed \(v_0\) at angle \(\theta_0\) above the horizontal. The initial components are \(v_{x0} = v_0\cos\theta_0\) and \(v_{y0} = v_0\sin\theta_0\). Integrating the constant accelerations (Lesson 3) gives
Projectile equations
\[ \begin{aligned} \colX{\text{Horizontal:}}\quad & v_x = v_{x0}, & x &= x_0 + v_{x0}\,t \\ \colY{\text{Vertical:}}\quad & v_y = v_{y0} - g t, & y &= y_0 + v_{y0}\,t - \tfrac12 g t^2 \\ & v_y^2 = v_{y0}^2 - 2g\,(y - y_0) & & \end{aligned} \]Launch and landing at the same height
For a launch from level ground that lands at the same height (\(y = y_0\)), three results come up so often that they are worth deriving once.
- Time to the top: \(v_y = 0\) when \(t = v_{y0}/g\). By symmetry the flight lasts twice as long: \(T = 2v_0\sin\theta_0/g\).
- Maximum height: from \(v_y^2 = v_{y0}^2 - 2g\,h\) with \(v_y = 0\): \(h = v_0^2\sin^2\theta_0/(2g)\).
- Range: \(R = v_{x0}\,T = v_0\cos\theta_0 \cdot 2v_0\sin\theta_0/g = v_0^2\sin 2\theta_0/g\).
Level ground only (landing height = launch height)
\[ T = \frac{2v_0\sin\theta_0}{g}, \qquad h_\text{max} = \frac{v_0^2\sin^2\theta_0}{2g}, \qquad R = \frac{v_0^2\sin 2\theta_0}{g} \]\(R\) is largest at \(\theta_0 = 45^\circ\), and complementary angles (\(\theta_0\) and \(90^\circ - \theta_0\)) give the same range.
Example 4.1 — A soccer kick
A ball is kicked from level ground at \(20\ \text{m/s}\), \(35^\circ\) above the horizontal. Ignoring air resistance, find the time of flight, the maximum height and the range.
Show solution
Components. \(v_{x0} = 20\cos 35^\circ = 16.38\ \text{m/s}\), \(v_{y0} = 20\sin 35^\circ = 11.47\ \text{m/s}\).
Time of flight. \(T = 2v_{y0}/g = 2(11.47)/9.81 = 2.339\ \text{s}\).
Maximum height. \(h_\text{max} = v_{y0}^2/(2g) = 11.47^2/19.62 = 6.707\ \text{m}\).
Range. \(R = v_{x0}T = 16.38 \times 2.339 = 38.32\ \text{m}\). Check: \(v_0^2\sin 70^\circ/g = 400(0.9397)/9.81 = 38.32\ \text{m}\). ✓
A strategy for any projectile problem
- Sketch the path. Put the origin at the launch point (or on the ground below it) and mark \(+x\) and \(+y\).
- List what you know for each direction: \(x_0, v_{x0}\); \(y_0, v_{y0}\), \(a_y = -g\). Note what the question asks for.
- Find the time of the event you care about (the top, the impact, reaching a wall). Usually one direction gives it: \(x\) when a horizontal distance is known, \(y\) when a height is known.
- Use that time in the other direction's equations.
- Check signs and units, and that the time is positive and sensible.
Example 4.2 — A stone thrown from a cliff
A stone is thrown horizontally at \(12\ \text{m/s}\) from the top of a cliff \(30\ \text{m}\) above the sea. Find how long it is in the air, how far from the foot of the cliff it lands, and its speed and direction at impact.
Show solution
Set up. Origin at the foot of the cliff, \(y\) up: \(y_0 = 30\ \text{m}\), \(v_{x0} = 12\ \text{m/s}\), \(v_{y0} = 0\).
Time (from \(y\)). Impact when \(y = 0\):
\[ 0 = 30 - \tfrac12(9.81)t^2 \quad\Rightarrow\quad t = \sqrt{\frac{60}{9.81}} = 2.473\ \text{s} \]Distance (from \(x\)). \(x = 12 \times 2.473 = 29.68\ \text{m}\).
Impact velocity. \(v_x = 12\ \text{m/s}\) (unchanged) and \(v_y = -9.81 \times 2.473 = -24.26\ \text{m/s}\):
\[ v = \sqrt{12^2 + 24.26^2} = 27.07\ \text{m/s}, \qquad \arctan\frac{24.26}{12} = 63.68^\circ \ \text{below the horizontal} \]Check. The fall time does not depend on the horizontal speed: a stone simply dropped from the cliff lands at the same instant.
The trajectory equation: aiming at a target
Eliminate \(t\) between \(x\) and \(y\). With the origin at the launch point, \(t = x/(v_0\cos\theta_0)\), and substituting into \(y(t)\) gives the path itself:
Trajectory (origin at the launch point)
\[ y = x\tan\theta_0 - \frac{g\,x^2}{2v_0^2\cos^2\theta_0} \]The path is a parabola, opening downward. Using \(1/\cos^2\theta_0 = 1 + \tan^2\theta_0\) turns it into a quadratic in \(\tan\theta_0\), which is how you find the launch angle that passes through a given point.
Example 4.3 — Aiming a fire hose
A firefighter holds a nozzle \(1\ \text{m}\) above the ground, \(15\ \text{m}\) from a building. Water leaves at \(20\ \text{m/s}\). At what angle must the jet be aimed to enter a window \(8\ \text{m}\) above the ground?
Show solution
Target relative to the nozzle. \(x = 15\ \text{m}\), \(y = 8 - 1 = 7\ \text{m}\). Let \(T = \tan\theta_0\).
Substitute into the trajectory equation with \(1/\cos^2\theta_0 = 1 + T^2\):
\[ 7 = 15T - \frac{9.81(15)^2}{2(20)^2}(1 + T^2) = 15T - 2.759(1 + T^2) \] \[ 2.759T^2 - 15T + 9.759 = 0 \quad\Rightarrow\quad T = \frac{15 \pm \sqrt{225 - 107.7}}{5.518} \]So \(T = 0.7556\) or \(T = 4.681\), giving \(\theta_0 = 37.08^\circ\) or \(\theta_0 = 77.94^\circ\).
Interpret. Both jets reach the window. The low one takes \(t = 15/(20\cos 37.08^\circ) = 0.940\ \text{s}\) and is still rising when it arrives (\(v_y = +2.84\ \text{m/s}\)); the high one takes \(3.59\ \text{s}\) and arrives falling steeply (\(v_y = -15.66\ \text{m/s}\)). The firefighter would choose the low, fast, direct jet. If the discriminant had been negative, no angle would reach the window at this speed.
Check your understanding
Key takeaways
- Projectile model: \(a_x = 0\), \(a_y = -g\). The horizontal and vertical motions are independent and share only \(t\).
- \(x = x_0 + v_0\cos\theta_0\,t\) and \(y = y_0 + v_0\sin\theta_0\,t - \tfrac12 g t^2\); at the top, \(v_y = 0\) but \(a_y = -g\).
- Level ground only: \(T = 2v_0\sin\theta_0/g\), \(h_\text{max} = v_0^2\sin^2\theta_0/(2g)\), \(R = v_0^2\sin 2\theta_0/g\).
- Strategy: find the time of the event from one direction, then use it in the other.
- The trajectory \(y = x\tan\theta_0 - g x^2/(2v_0^2\cos^2\theta_0)\) is a parabola; solving it for \(\tan\theta_0\) aims at a target.
- Next: Lesson 5 introduces path coordinates, which follow the particle instead of fixed axes.