Coaching Centre
Deriving the projectile equations
Deriving time of flight, maximum height and range from the HSC equations, why 45° gives the greatest range, complementary launch angles, launching from a height, final velocity, and how air resistance and spin change real trajectories.
An artillery officer in 1700 had no calculator, but did have a rule of thumb: elevate the gun to 45° for maximum range. On a flat field with no air, that rule is exactly right, and you can prove it in four lines of algebra. On a real battlefield it is wrong, and knowing why is just as useful.
The HSC formulae sheet doesn't give you formulas for range or maximum height. It gives you the equations of motion, and the syllabus expects you to derive the relationships yourself. Each derivation below starts from the sheet.
Time to the top and time of flight
For a projectile launched from ground level at speed $u$ and angle $\theta$, $u_y = u\sin\theta$. At the top, $v_y = 0$:
$0 = u\sin\theta - gt \quad\Rightarrow\quad t_{\text{top}} = \dfrac{u\sin\theta}{g}$
If it lands at the same height it was launched from, the flight is symmetrical, so the total time of flight is double this. You can also get it from $\Delta y = 0$:
$0 = (u\sin\theta)t - \tfrac{1}{2}gt^2 \quad\Rightarrow\quad t_{\text{flight}} = \dfrac{2u\sin\theta}{g}$
Maximum height
Use $v_y^2 = u_y^2 + 2a_y\Delta y$ with $v_y = 0$ and $a_y = -g$:
$0 = u^2\sin^2\theta - 2gH \quad\Rightarrow\quad H = \dfrac{u^2\sin^2\theta}{2g}$
Maximum height depends only on the vertical component of the launch velocity. Doubling the launch speed makes it four times higher.
Range, and why 45° wins
Horizontal range is horizontal velocity × time of flight:
$R = (u\cos\theta)\left(\dfrac{2u\sin\theta}{g}\right) = \dfrac{u^2 \cdot 2\sin\theta\cos\theta}{g} = \dfrac{u^2\sin 2\theta}{g}$
For a fixed launch speed, $R$ is largest when $\sin 2\theta = 1$, so $2\theta = 90°$ and $\theta = 45°$. The maximum range is then $u^2/g$.
Because $\sin 2\theta = \sin(180° - 2\theta)$, angles that add to 90° give the same range. A launch at 30° and one at 60° land in the same spot; the 60° shot goes higher and takes longer. These are called complementary angles.
Launching from a height
When the launch point is above the ground, the simple formulas above no longer apply, because the flight isn't symmetrical. Go back to the equations of motion. The vertical displacement equation becomes a quadratic in $t$; take the positive root.
A ball is thrown from a cliff 20 m high at 15 m/s, 35° above the horizontal. Find the time of flight, the range, the maximum height above the ground and the impact velocity.
That last check is worth remembering. With no air resistance, mechanical energy is conserved, so the final speed of a projectile is $v = \sqrt{u^2 + 2gh}$ whatever the launch angle. The angle only changes the direction of impact. And from a height, the best angle for range is less than 45°: a flatter launch spends the extra fall time moving forward.
- "Bin on the ground": find an angle that hits the bin (try around 23° to 24°). Then find a second, completely different angle that also works. Do your two angles add to 90°?
- Press Show both solutions and compare them with the angles you found.
- "Over the wall": both solutions reach the target distance. Why does only one of them work?
- Switch to Range against angle. From the ground, where is the peak? Now launch from 10 m (at 20 m/s the best angle is 39.3°, for 49.8 m) and then from 20 m. What happens to the best angle as the launch gets higher?
What the model leaves out
Air resistance acts opposite to the velocity, so it slows both components. The real path is lower and shorter, it is no longer symmetrical (the descent is steeper than the climb), and the best launch angle drops below 45°. For a thrown cricket ball it is around 40°; for a long jumper, who can't run fast and jump steeply, the take-off angle is only about 20°.
Spin matters too. A golf ball's backspin creates lift (the Magnus effect), so drives are launched at only 10° to 15°. A topspin forehand in tennis dips sharply for the same reason in reverse. None of this is in the HSC model, but examiners do ask you to evaluate the model, and these are the reasons it falls short.
Test yourself
Two projectiles are launched from level ground at the same speed, at 25° and 65°. Which statement is correct?
25° + 65° = 90°, so $\sin 2\theta$ is equal; the steeper launch has the larger $u_y$, so it flies higher and longer.
The launch speed of a projectile on level ground is doubled, with the angle unchanged. The range
$R \propto u^2$.
A ball is launched at 20 m/s at 60° above level ground. Its maximum height is closest to
$H = \frac{(20\sin60°)^2}{2(9.8)} = \frac{300}{19.6} = 15.3$ m.
Three balls are thrown from the same cliff at the same speed: one horizontally, one 30° up and one 30° down. Ignoring air resistance, which hits the water with the greatest speed?
Energy is conserved, so $v = \sqrt{u^2 + 2gh}$ for all three. Only the directions differ.
Label the quantities for a launch from a cliff.
Order these launches (all at 20 m/s from level ground) from shortest to longest range.
From level ground, the range is greatest at a launch angle of . Launch angles that add to 90° are called angles and give the same range. Maximum height depends only on the component of the launch velocity. When launched from a height, the time of flight is found by solving a equation. Air resistance makes the best launch angle than 45°.
Starting from the equations of motion, derive an expression for the horizontal range of a projectile launched from level ground with speed $u$ at angle $\theta$. Use it to show that launch angles of 30° and 60° give the same range, and explain why 45° gives the maximum range.
Assume constant g, no air resistance and level ground; take up as positive. Resolve: $u_x = u\cos\theta$, $u_y = u\sin\theta$.
Vertically, the projectile lands when $\Delta y = 0$: $0 = u\sin\theta\, t - \tfrac12 g t^2$, so $t = \frac{2u\sin\theta}{g}$ (ignoring $t = 0$).
Horizontally, $R = u_x t = u\cos\theta \cdot \frac{2u\sin\theta}{g} = \frac{u^2 \cdot 2\sin\theta\cos\theta}{g} = \frac{u^2\sin2\theta}{g}$.
At 30°, $\sin 2\theta = \sin60° = 0.866$; at 60°, $\sin2\theta = \sin120° = 0.866$. The values are equal, so the ranges are equal.
For a given $u$, $R$ is largest when $\sin2\theta$ is largest. Its maximum value is 1, when $2\theta = 90°$, i.e. $\theta = 45°$, giving $R_{\max} = u^2/g$.