Projectile motion: range, maximum height, formulas

In Year 4 of the liceo classico (Italian classics-focused high school), projectile motion usually arrives in physics in October or shortly after, often together with circular motion. Many teachers cover it quickly because the programme is long: the student then meets it again in a test, and the test that includes it usually falls six to eight weeks from now, between late November and December. It is worth understanding it now, calmly, because once the basic idea is clear the formulas rebuild themselves.

Here you will find the idea (two independent motions), the formulas for range, maximum height and time of flight, the mistakes I see most often and three worked exercises. For circular motion and for work and energy I have written a separate article; for the basics of s-t and v-t graphs, the one on kinematics; for oscillations, the one on simple harmonic motion. I do not repeat them here.

The idea: two independent motions

A stone thrown horizontally off a table does not fall straight down: it traces a curve. Galileo explained it like this: the motion is made up of two motions that do not influence each other.

• Horizontally no force acts (air is neglected): the motion is uniform motion in a straight line, with constant velocity $v_x$.
• Vertically only gravity acts: the motion is uniformly accelerated, with acceleration $g=9.8{\text{m/s}}^2$ downwards.

Time is the only link between the two. For this reason, in problems, you solve the vertical direction first (which gives you the time) and then use that time horizontally. One surprising consequence: a projectile fired horizontally and one dropped from the same height at the same instant hit the ground together.

All the formulas that follow hold in the model without air resistance, which is the one school asks for.

Horizontal launch

You start from height $h$ with horizontal velocity $v_0$. The equations of motion are:

The body reaches the ground when $y=0$, so the time of flight depends only on the height:

The range (horizontal distance covered) is $x=v_0{t}_v$. The velocity at any instant has two components: $v_x=v_0$, always the same, and $v_y=gt$, which grows. The magnitude is $v=\sqrt{v_x^2+v_y^2}$.

Angled launch

You launch from the ground with velocity $v_0$ at an angle $\theta$ to the horizontal. You resolve the vector into components:

The equations of motion become:

Three quantities to know, valid when you start and land at the same height. The time of flight is twice the time of ascent, and the time of ascent is when $v_y$ becomes zero, that is $t=v_0\sin \theta /g$:

The maximum height is reached exactly halfway through the flight, at the point where the body has only the horizontal velocity:

The range is obtained by multiplying the horizontal velocity by the time of flight and using $2\sin \theta \cos \theta =\sin 2\theta$:

From here follow the two most frequent theory questions. The range is maximum when $\sin 2\theta =1$, that is for $\theta =45°$, and it equals $v_0^2/g$. And two complementary angles ($\theta$ and $90°-\theta$) give the same range, because $\sin 2\theta =\sin (180°-2\theta )$: the lower launch arrives sooner, the higher one climbs more and stays in the air longer.

If the problem asks for the shape of the trajectory, eliminating time gives a parabola:

At the highest point the velocity is not zero: the horizontal component $v_0\cos \theta$ is still there. This is one of the most common traps.

The mistakes that come back every year

Forgetting to resolve the velocity. With an angled launch, $v_0$ is neither $v_x$ nor $v_y$. If you use $v_0$ in place of $v_0\sin \theta$ in the ascent formula, you get a height that is too large.

Saying that the velocity is zero at the highest point. Only the vertical component is zero. The magnitude of the velocity there is $v_0\cos \theta$.

A non-zero horizontal acceleration. In the basic model $a_x=0$ always: if you put $g$ horizontally as well you get meaningless numbers.

Calculator in the wrong mode. If the angle is in degrees, the calculator must be in degrees: $\sin 30°=0.5$, whereas $\sin (30\text{rad})=-0.988$. Before calculating, check that $\sin 30°$ gives you $0.5$.

Using the range formulas when you do not start and land at the same height. $R=v_0^2\sin 2\theta /g$ and $t_v=2v_0\sin \theta /g$ hold only in that case. If you launch from a balcony, you need to solve $y(t)=0$ for the time.

Confusing maximum height and starting height. In a launch from the ground $h_{max}$ is the height above the ground; if you start from a height $h_0$, the maximum height is $h_0+h_{max}$.

Three worked exercises

1. Horizontal launch from a cliff

A stone is thrown horizontally at $v_0=15$ m/s from a height $h=45$ m ($g=9.8$ m/s²). Calculate the time of flight, the range and the velocity (magnitude and angle) at impact.

The time depends only on the height:

The range:

At impact the vertical component is $v_y=g{t}_v=9.8\cdot 3.03\approx 29.7$ m/s, and the horizontal one is still $15$ m/s:

The angle below the horizontal: $\tan \alpha =29.7/15\approx 1.98$, so $\alpha \approx 63°$. Check: if the stone fell straight down from the same height it would arrive with $v_y=29.7$ m/s after $t=3.03$ s, so the time is right.

2. Angled launch: time, height, range

A ball is thrown from the ground at $v_0=20$ m/s at an angle of $30°$. Find the time of flight, maximum height and range. Then: what range would you get with $60°$?

Components: $v_{0x}=20\cos 30°\approx 17.3$ m/s and $v_{0y}=20\sin 30°=10$ m/s.

Time of flight:

Maximum height:

Range:

Check with the direct formula: $R=20^2\sin 60°/9.8=400\cdot 0.866/9.8\approx 35.4$ m, the same.

With $60°$ the components swap ($v_{0x}=10$ m/s, $v_{0y}\approx 17.3$ m/s): the time of flight becomes $2\cdot 17.3/9.8\approx 3.53$ s and the range $10\cdot 3.53\approx 35.4$ m, identical. What changes is the maximum height, which rises to about $15.3$ m.

3. Which angle hits a target?

A cannon fires with $v_0=25$ m/s. The target is $40$ m away on the same level. What is the maximum possible range? At which angles is the target hit?

The maximum range is at $45°$:

The target at $40$ m is reachable, because $40<63.8$. We set $R=40$ m:

So $2\theta \approx 38.8°$ or $2\theta \approx 180°-38.8°=141.2°$, that is:

The two angles are complementary ($19.4°+70.6°=90°$), as expected. The flat shot arrives in about $1.7$ s, the high-arc one in about $4.8$ s: same range, very different times.

How to prepare over the coming weeks

You have time, so do a little, often:

1. Learn the procedure more than the formulas: resolve the velocity, solve the vertical direction for the time, use the time horizontally.
2. Redo the three exercises without looking, changing the angle and the speed.
3. Draw the trajectory with the vectors $v_x$ and $v_y$ at four points: start, ascent, top, landing.
4. When the topic is closed in class, practise against the clock with the simulator for high school tests. There is no dedicated test on this topic yet, but the physics questions tell you where you stand with the rest of the programme.

More material on the subject is on the physics page.

How we work in lessons

I give one-to-one tutoring online. With students who come to me on this topic I start from a drawing with the two directions separated, because once you look at the motion as two motions the formula is no longer something to learn by heart. Then we do exercises with your own numbers, especially the ones where the angle or the unit of measurement hides a mistake.

If you want a hand

If the test is getting close and you want to see where you stand, you can book a free introductory call: we look together at where the gaps are and whether it makes sense to work on them.