Interactive Lecture
Projectile Motion Under Gravity
by Dr Arnaldo Delli Carri
Projectile motion is a form of motion where an object moves in a parabolic path under the influence of gravity. This kind of motion is a combination of:
A projectile is any object that is thrown or launched into the air and is subject only to:
For an object launched with an initial velocity \(v_0\) at an angle \(\theta\) with the horizontal:
Horizontal position: \(x = (v_0 \cos \theta) \cdot t\)
Vertical position: \(y = (v_0 \sin \theta) \cdot t - \frac{1}{2}gt^2\)
Horizontal velocity: \(v_x = v_0 \cos \theta\) (constant)
Vertical velocity: \(v_y = v_0 \sin \theta - gt\)
The path of a projectile is always a parabola when:
The maximum height reached by a projectile is given by:
\(h_{max} = \frac{(v_0 \sin \theta)^2}{2g}\)
The horizontal distance travelled before returning to the launch height is:
\(R = \frac{v_0^2 \sin(2\theta)}{g}\)
The maximum range occurs at \(\theta = 45°\)
For a given launch speed, the range is greatest at 45° — assuming the projectile lands at the same height it was launched from.
The total time the projectile is in the air:
\(t_{flight} = \frac{2v_0 \sin \theta}{g}\)
Use your mouse to set the angle and initial velocity of the projectile. Click to launch!
The understanding of projectile motion has evolved significantly throughout history:
Aristotle believed that objects moved only as long as a force acted upon them. He proposed that projectiles continued moving after leaving the thrower's hand because the air behind them pushed them forward—a theory that persisted for nearly 2,000 years.
Scholars like John Philoponus and Jean Buridan developed the "impetus" theory, suggesting that objects were imbued with a temporary force or "impetus" that gradually dissipated.
Galileo Galilei revolutionised our understanding by demonstrating that projectile motion could be separated into two independent components: horizontal motion at constant velocity and vertical motion with constant acceleration. He was the first to prove mathematically that the path of a projectile is a parabola.
Galileo's experiments with inclined planes helped him understand acceleration due to gravity, laying the groundwork for Newton's laws of motion, which would give the first complete mathematical account of projectile motion half a century later.
Isaac Newton's laws of motion and universal gravitation provided the complete mathematical foundation for understanding projectile motion that we still use today.
Air resistance reshapes projectile motion once speeds are high enough for drag to bite. When it is included:
The drag force is proportional to the square of velocity:
\(F_d = -\frac{1}{2}C_d \rho A v^2\)
Where:
With air resistance:
The acceleration due to gravity varies on different celestial bodies, affecting projectile motion:
For the same initial velocity and angle, a projectile on the Moon travels about 6 times farther than on Earth.
When a projectile is launched from height \(h\) above the ground:
Time of flight: \(t_{flight} = \frac{v_0\sin\theta + \sqrt{(v_0\sin\theta)^2 + 2gh}}{g}\)
Range: \(R = v_0\cos\theta \cdot t_{flight}\)
Projectile motion appears in more engineering contexts than you might expect:
| Concept | Equation | Description |
|---|---|---|
| Horizontal Position | \(x = (v_0 \cos \theta) \cdot t\) | \(v_0\) = initial velocity \(\theta\) = launch angle \(t\) = time |
| Vertical Position | \(y = (v_0 \sin \theta) \cdot t - \frac{1}{2}gt^2\) | \(g\) = gravitational acceleration |
| Horizontal Velocity | \(v_x = v_0 \cos \theta\) | Constant throughout motion |
| Vertical Velocity | \(v_y = v_0 \sin \theta - gt\) | Decreases linearly with time |
| Maximum Height | \(h_{max} = \frac{(v_0 \sin \theta)^2}{2g}\) | Height at which \(v_y = 0\) |
| Range (same height) | \(R = \frac{v_0^2 \sin(2\theta)}{g}\) | Horizontal distance travelled |
| Time of Flight | \(t_{flight} = \frac{2v_0 \sin \theta}{g}\) | Total time in air |
| Time to Maximum Height | \(t_{max} = \frac{v_0 \sin \theta}{g}\) | Time to reach highest point |
Before you move on, make sure you can state these five points confidently:
These principles form the foundation for understanding more complex motion in engineering, sports, and natural phenomena.
A ball is thrown with an initial velocity of 15 m/s at an angle of 30° with the horizontal. Calculate:
Show Solution
Given: \(v_0 = 15 \text{ m/s}, \theta = 30°, g = 9.81 \text{ m/s}^2\)
a) Maximum height:
\(h_{max} = \frac{(v_0 \sin \theta)^2}{2g} = \frac{(15 \sin 30°)^2}{2 \times 9.81} = \frac{(15 \times 0.5)^2}{19.62} = \frac{56.25}{19.62} = 2.87 \text{ m}\)
b) Horizontal range:
\(R = \frac{v_0^2 \sin(2\theta)}{g} = \frac{15^2 \sin(60°)}{9.81} = \frac{225 \times 0.866}{9.81} = 19.86 \text{ m}\)
c) Time of flight:
\(t_{flight} = \frac{2v_0 \sin \theta}{g} = \frac{2 \times 15 \times 0.5}{9.81} = \frac{15}{9.81} = 1.53 \text{ s}\)
Answer: (a) \(h_{max} = 2.87\,\text{m}\); (b) \(R = 19.86\,\text{m}\); (c) \(t_{flight} = 1.53\,\text{s}\)
A football player wants to kick a ball to reach a maximum height of 20 metres. If the ball is kicked at an angle of 60° to the horizontal, what initial velocity is required?
Show Solution
Given: \(h_{max} = 20 \text{ m}, \theta = 60°, g = 9.81 \text{ m/s}^2\)
Using the equation for maximum height:
\(h_{max} = \frac{(v_0 \sin \theta)^2}{2g}\)
Rearranging to solve for \(v_0\):
\(v_0 = \sqrt{\frac{2g \times h_{max}}{\sin^2 \theta}} = \sqrt{\frac{2 \times 9.81 \times 20}{\sin^2 60°}} = \sqrt{\frac{392.4}{0.75}} = \sqrt{523.2} = 22.87 \text{ m/s}\)
Answer: \(v_0 = 22.87\,\text{m/s}\)
A cannon is positioned 20 metres below a target on a cliff. If the cannon can fire a projectile at 50 m/s, at what angle should it be fired to hit the target, which is 100 metres away horizontally?
Show Solution
Given: \(v_0 = 50 \text{ m/s}, \Delta x = 100 \text{ m}, \Delta y = 20 \text{ m}, g = 9.81 \text{ m/s}^2\)
Using the parametric equations:
\(x = (v_0 \cos \theta) \cdot t\)
\(y = (v_0 \sin \theta) \cdot t - \frac{1}{2}gt^2\)
First, find the flight time from the horizontal equation:
\(t = \frac{\Delta x}{v_0 \cos \theta} = \frac{100}{50 \cos \theta} = \frac{2}{\cos \theta}\)
Substituting into the vertical equation and setting \(y = 20\):
\[20 = 50 \sin\theta \cdot \frac{2}{\cos\theta} - \frac{1}{2}(9.81)\left(\frac{2}{\cos\theta}\right)^2\]
\[20 = 100\tan\theta - \frac{1}{2}(9.81)(4\sec^2\theta)\]
\[20 = 100\tan\theta - 19.62\sec^2\theta\]
Replacing \(\sec^2\theta = 1 + \tan^2\theta\) and letting \(u = \tan\theta\):
\[20 = 100u - 19.62(1 + u^2)\]
\[19.62u^2 - 100u + 39.62 = 0\]
Using the quadratic formula:
\[u = \frac{100 \pm \sqrt{10000 - 4(19.62)(39.62)}}{2(19.62)} = \frac{100 \pm \sqrt{6891}}{39.24} = \frac{100 \pm 83.01}{39.24}\]
Two solutions exist (low and high trajectory):
\[u_1 = \frac{100 - 83.01}{39.24} = 0.433 \implies \theta_1 = \arctan(0.433) \approx 23.4°\]
\[u_2 = \frac{100 + 83.01}{39.24} = 4.664 \implies \theta_2 = \arctan(4.664) \approx 77.9°\]
The practical (low-trajectory) solution is \(\theta \approx 23.4°\). The high-trajectory solution of 77.9° also works but would be impractical for a cannon.
Answer: \(\theta \approx 23.4°\) (low trajectory) or \(\theta \approx 77.9°\) (high trajectory)
Verification (\(\theta = 23.4°\)): \(t = 2/\cos(23.4°) = 2.18\) s; \(y = 50\sin(23.4°)\times 2.18 - \frac{1}{2}(9.81)(2.18)^2 = 43.27 - 23.31 \approx 20\) m ✓