This Year 13 projectiles and variable acceleration revision notes is written for pupils aged 17-18 and follows the National Curriculum. In short, it shows you how to model projectile motion and use calculus for variable acceleration. Before you start, it helps to revise conditional probability and the Normal distribution and hypothesis tests: correlation and Normal mean, because this lesson builds on them.
Year 13 projectiles and variable acceleration revision notes: what you will learn
A projectile is a particle moving freely under gravity alone. This chapter models projectile motion, then uses calculus for motion with variable acceleration.
Starter activity
Throw a ball to a friend and notice its path. At which point is it moving most slowly? Explain why a ball thrown at \(45^\circ\) travels further than one thrown at \(15^\circ\) or \(75^\circ\). Now, think about a car driving on a motorway: why does the acceleration change with time as the driver presses the accelerator, and what graph would you draw to show this?
Modelling projectile motion
Definition: projectile model
In the usual model we ignore air resistance, treat the object as a particle, assume gravity is constant, acts downwards with \(g=9.8\ \mathrm{m\,s^{-2}}\), and ignore the wind. Resolve the motion into horizontal and vertical components. Horizontally there is no acceleration; vertically the acceleration is \(-g\), so the suvat equations apply. Projected with speed \(u\) at angle \(\theta\) above the horizontal, the initial components are \(u\cos\theta\) and \(u\sin\theta\).
The equations of motion
Rule: projectile equations
\(x=ut\cos\theta\), \(y=ut\sin\theta-\tfrac12gt^2\), \(v_x=u\cos\theta\), \(v_y=u\sin\theta-gt\). The speed is \(\sqrt{v_x^2+v_y^2}\) at angle \(\tan^{-1}\dfrac{|v_y|}{v_x}\) to the horizontal. Eliminating \(t\): \(y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}\), a downward parabola.
Key fact: top of the flight and range
At the highest point \(v_y=0\): time \(\dfrac{u\sin\theta}{g}\), height \(H=\dfrac{u^2\sin^2\theta}{2g}\), and the particle moves horizontally at \(u\cos\theta\). Landing at the launch height: \(T=\dfrac{2u\sin\theta}{g}\) and range \(R=\dfrac{u^2\sin2\theta}{g}\), greatest at \(\theta=45^\circ\).
Worked example 1
A particle is projected at \(u=20\) and \(\theta=30^\circ\). Find \(T\), \(R\) and \(H\).
\(T=2.04\) s, \(R=35.3\) m and \(H=5.10\) m. For \(u=25\) with \(\tan\theta=\tfrac34\), the height at \(x=40\) m is \(30-19.6=10.4\) m.
Method: projectile questions
- Sketch the launch point, initial velocity and angle.
- Write two columns, "horizontal" and "vertical", and fill in what is known. Take upwards as positive so \(g\) appears as \(-9.8\).
- The shared quantity is the time \(t\): find it from one column and use it in the other.
- For speed or direction at a point, find both components and combine them with Pythagoras and a tangent. "At the highest point": \(v_y=0\).
- Give 3 significant figures, with units.
Worked example 2
A stone is thrown horizontally at \(12\ \mathrm{m\,s^{-1}}\) from a cliff 45 m high. Find the distance and the landing speed.
\(45=\tfrac12(9.8)t^2\), so \(t=3.03\) s and the horizontal distance is \(12\times3.03=36.4\) m. At landing \(v_y=29.7\), so the speed is \(\sqrt{12^2+29.7^2}=32.0\ \mathrm{m\,s^{-1}}\) at \(68.0^\circ\) below the horizontal.
Worked example 3
A ball hit at 30 m/s lands 60 m away on level ground. Find the angle.
\(\sin2\theta=\dfrac{gR}{u^2}=\dfrac{588}{900}=0.6533\), so \(2\theta=40.8^\circ\) or \(139.2^\circ\), and \(\theta=20.4^\circ\) or \(69.6^\circ\). The two angles add to \(90^\circ\) and give the same range.
Worked example 4
A ball at 20 m/s and \(35^\circ\) meets a wall 30 m away. Find its height there, and its speed at \(y=3\) m for \(u=20\).
\(t=\dfrac{30}{20\cos35^\circ}=1.83\) s, \(y=11.47(1.83)-4.9(1.83)^2=4.58\) m. Using \(v^2=u^2-2gy\): \(v=\sqrt{400-58.8}=18.5\ \mathrm{m\,s^{-1}}\).
Watch out
Do not swap sine and cosine when resolving. \(v_y\) changes sign at the top, but only the vertical component is zero there: the speed at the top is \(u\cos\theta\), not 0.
Variable acceleration with calculus
Rule: calculus in kinematics
When acceleration is not constant the suvat equations do not apply. Use \(v=\dfrac{ds}{dt}\), \(a=\dfrac{dv}{dt}\), and in reverse \(v=\int a\,dt\), \(s=\int v\,dt\). Use the initial conditions to find each constant.
Worked example 5
\(a=6t-4\), with \(v=2\) and \(s=0\) at \(t=0\). Find \(v\) and \(s\).
\(v=3t^2-4t+2\) and \(s=t^3-2t^2+2t\).
Worked example 6
\(v=3t^2-2t+1\). Find \(a\) at \(t=2\) and the displacement from \(t=0\) to \(2\). Then for \(v=4t-t^2\) find when it is at rest, the maximum speed and the distance in the first 4 s.
\(a=6t-2=10\) at \(t=2\); displacement \(\int_0^2v\,dt=8-4+2=6\) m. For \(v=4t-t^2\): at rest at \(t=0\) and \(4\); \(\dfrac{dv}{dt}=4-2t=0\) at \(t=2\), maximum \(v=4\); distance \(\int_0^4(4t-t^2)\,dt=\tfrac{32}3\) m.
Displacement versus distance
Definition: displacement and distance
\(\int v\,dt\) gives displacement, which is signed. The total distance needs \(v=0\) found first, then each part integrated separately and taken positive.
Worked example 7
\(v=t^2-4t+3\) from \(t=0\) to \(3\). Find the displacement and the distance.
\(v=(t-1)(t-3)\). The displacement is \(0\), but the particle moves forward \(\tfrac43\) m then back \(\tfrac43\) m, so the distance is \(\tfrac83\) m.
Worked example 8
\(\mathbf r=t^2\mathbf i+(2t^3-3t)\mathbf j\). Find \(\mathbf v\), \(\mathbf a\) and the speed at \(t=1\).
\(\mathbf v=2t\mathbf i+(6t^2-3)\mathbf j\), \(\mathbf a=2\mathbf i+12t\mathbf j\). At \(t=1\): \(\mathbf v=2\mathbf i+3\mathbf j\), speed \(\sqrt{13}=3.61\ \mathrm{m\,s^{-1}}\).
Commenting on the model
Note: limitations
Air resistance would reduce range and height; the object is a particle, so spin and size are ignored; \(g\) is constant over small heights; wind is ignored. Say what effect the improvement would have, not just that it exists.
Watch out
Use initial conditions to find constants. Do not use \(\int v\,dt\) for the total distance when the particle changes direction.
Note: the path of a projectile
Worked example 9: Speed at a given height
A ball is projected at 20 m s\(^{-1}\) at \(35^\circ\). Find its speed when it is 5 m above the launch point.
Use conservation of the horizontal component and \(v_y^2=u_y^2-2gy\). \(u_x=20\cos35^\circ=16.38\), \(u_y=11.47\). At \(y=5\): \(v_y^2=131.6-98=33.6\), so \(v_y=\pm5.80\). Speed \(=\sqrt{16.38^2+33.6}=\sqrt{268.3+33.6}=17.4\) m s\(^{-1}\). The same speed applies on the way up and on the way down at this height.
Worked example 10: Does the ball hit the target?
A stone is thrown at 12 m s\(^{-1}\) at \(30^\circ\) from ground level towards a wall 10 m away. Find the height at which it hits the wall.
\(t=\dfrac{10}{12\cos30^\circ}=0.962\) s. \(y=12\sin30^\circ\times0.962-4.9\times0.962^2=5.77-4.53=1.24\) m. Check that the stone has not already landed: the time of flight is \(\dfrac{2\times6}{9.8}=1.22\) s, which is longer than 0.962 s, so it reaches the wall at 1.24 m and is going downwards (since \(0.962>6/9.8=0.61\) s).
Worked example 11: Integrating with a limit
A particle has acceleration \(a=12t-6\) m s\(^{-2}\) and starts at the origin with velocity \(-2\) m s\(^{-1}\). Find the time when it returns to the origin.
\(v=6t^2-6t-2\) (with \(C=-2\)). \(s=2t^3-3t^2-2t\) (with \(s=0\) at \(t=0\)). Solve \(s=0\): \(t(2t^2-3t-2)=0\), so \(t(2t+1)(t-2)=0\) and \(t=2\) s (rejecting \(t=0\) and \(t=-\tfrac12\)).
Worked example 12: Maximum range from a height
A ball is thrown from a roof, 20 m above level ground, at 15 m s\(^{-1}\) horizontally. How far from the building does it land?
\(20=4.9t^2\), so \(t=2.02\) s. Distance \(=15\times2.02=30.3\) m.
Worked example 13: Finding when a particle is at its maximum displacement
A particle moves with \(v=6t-3t^2\) m s\(^{-1}\), starting at the origin. Find the greatest displacement from the origin and when it occurs.
\(v=3t(2-t)=0\) at \(t=0\) and \(t=2\). The displacement is \(s=3t^2-t^3\), so \(s(2)=12-8=4\) m. After \(t=2\) the velocity is negative, so the particle returns towards the origin and reaches it again when \(3t^2-t^3=0\), i.e. \(t=3\) s. The greatest displacement is 4 m at \(t=2\) s.
Worked example 14: Time and height from the quadratic in t
A stone is thrown vertically upwards from the edge of a cliff at 14 m/s. The cliff top is 49 m above the sea. Find the time to reach the sea.
Take upwards as positive: \(-49=14t-4.9t^2\), so \(4.9t^2-14t-49=0\), or \(t^2-\tfrac{20}{7}t-10=0\). Using the formula, \(t=\dfrac{14+\sqrt{196+960.4}}{9.8}=\dfrac{14+34.0}{9.8}=4.90\) s. The negative root is rejected because the time must be positive. The stone first rises for \(\dfrac{14}{9.8}=1.43\) s.
Worked example 15: Modelling remarks on a projectile
List three assumptions in the basic projectile model and the effect of removing each one for a tennis ball.
(1) No air resistance: with air resistance the ball would travel less far and land at a steeper angle. (2) The ball is a particle, so size and spin are ignored: topspin makes a tennis ball dip faster, backspin makes it float. (3) Constant \(g\) and no wind: a following wind would increase the range, a head wind would reduce it. Each comment says what effect the change would have, which is what examiners look for.
Rule: working with the suvat equations in each direction
For constant acceleration use \(v=u+at\), \(s=ut+\tfrac12at^2\), \(v^2=u^2+2as\) and \(s=\tfrac12(u+v)t\) in each direction separately. The same time \(t\) appears in both. Horizontally \(a=0\), so \(s=ut\). Vertically \(a=-g\), so \(s=ut-\tfrac12gt^2\). For a particle that returns to its launch level, the time up equals the time down and the landing speed equals the launch speed.
Method: projectile or variable acceleration?
- Is the acceleration constant? For a projectile it is \(0\) horizontally and \(-g\) vertically, so use suvat in each direction separately.
- If the acceleration is given as a function of \(t\), integrate to find \(v\) and \(s\), using initial conditions to find each constant.
- If a position or velocity is given, differentiate to find \(v\) and \(a\).
- For distance rather than displacement, find where \(v=0\) and integrate between those times, taking each part as positive.
Watch out: signs and speed at the top
Choose upwards as positive and use \(-9.8\) for acceleration. At the top of a projectile path the vertical component is zero but the horizontal component is not, so the speed is \(u\cos\theta\). Also, the magnitude of the displacement is not the distance travelled when the particle turns round: check where \(v=0\).
Did you know?
Galileo Galilei showed around 1600 that a projectile follows a parabola, by treating its horizontal and vertical motion separately, an idea that was hard to accept at the time because many people believed in an older theory that a cannonball moved in a straight line until it ran out of impetus. Today the same equations are used to design sports equipment, fountains and the paths of spacecraft after engine cut-off.
Otto’s tip
Draw the initial velocity triangle with the angle marked before you write any equation.
How to use this Year 13 projectiles and variable acceleration revision notes
First, read the key points so that you know where the Year 13 projectiles and variable acceleration revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the projectiles and variable acceleration worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the projectiles and variable acceleration test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 13 projectiles and variable acceleration revision notes gives you everything you need for the chapter.
Similar pages
WorksheetProjectiles and variable acceleration – Year 13 A level maths worksheet with answers to download as PDF
AnswersProjectiles and variable acceleration – Year 13 A level maths worksheet answers to download as PDF
Multiple choice questionsProjectiles and variable acceleration – Year 13 A level maths multiple choice questions to download as PDF
QuizProjectiles and variable acceleration – Year 13 A level maths quiz to download as PDF
TestProjectiles and variable acceleration – Year 13 A level maths test to download as PDF
Test answersProjectiles and variable acceleration – Year 13 A level maths test answers to download as PDF
StartersProjectiles and variable acceleration – Year 13 A level maths starters to download as PDFOther chapters in this level
LessonProof by contradiction – Year 13 A level maths revision notes to download as PDF
LessonFunctions, modulus and partial fractions – Year 13 A level maths revision notes to download as PDF
LessonBinomial expansion for rational n and series modelling – Year 13 A level maths revision notes to download as PDF
LessonFurther trigonometry – Year 13 A level maths revision notes to download as PDF
LessonParametric equations – Year 13 A level maths revision notes to download as PDF
LessonDifferentiation: chain, product and quotient rules – Year 13 A level maths revision notes to download as PDF
LessonIntegration techniques and differential equations – Year 13 A level maths revision notes to download as PDF
LessonNumerical methods – Year 13 A level maths revision notes to download as PDF