Before stepping up to resisted projectile motion, lock in the Extension 1 baseline: in the absence of air resistance, the horizontal velocity $u\cos\theta$ stays constant, while the vertical velocity changes uniformly as $u\sin\theta - gt$. Range, maximum height, and time of flight all follow from those two facts. This lesson is a fast review, get fluent here so you can focus on the drag term next.
+5 XP warm-up A projectile is launched at angle with speed (no air resistance).
A projectile is launched at angle $\theta$ with speed $u$ (no air resistance). Before checking what is the horizontal acceleration, and what is the vertical acceleration? At the highest point of the trajectory, what is the value of $\dot{y}$? Of $\dot{x}$?
+5 XP to read Every projectile question (without resistance) rewards two habits: resolve the initial velocity into horizontal...
Every projectile question (without resistance) rewards two habits: resolve the initial velocity into horizontal and vertical components $u\cos\theta$ and $u\sin\theta$, then treat each axis independently uniform velocity horizontally, uniform acceleration $-g$ vertically. Coupling between the axes only comes back when you set $y = 0$ to find when the projectile lands.
The resolve-integrate-couple workflow: (1) resolve $\vec{u}$ into $(u\cos\theta, u\sin\theta)$, (2) integrate the constant accelerations $(0, -g)$ twice for $x(t)$ and $y(t)$, (3) couple the axes only via $t$ (e.g. set $y = 0$ to find the landing time, then read off $x$).
$\dot{x} = u\cos\theta$ · $\dot{y} = u\sin\theta - gt$ · $x = u\cos\theta\cdot t$ · $y = u\sin\theta\cdot t - \tfrac12 gt^2$
Take horizontal, vertical (up positive).
Take $x$ horizontal, $y$ vertical (up positive). With no air resistance, the only force is gravity, so
Time of flight on level ground: set $y = 0$ and $t \neq 0$: $u\sin\theta = \tfrac12 gt$, so $T = \dfrac{2u\sin\theta}{g}$.
Range: $R = u\cos\theta \cdot T = \dfrac{2u^2 \sin\theta\cos\theta}{g} = \dfrac{u^2 \sin 2\theta}{g}$. Maximum when $\sin 2\theta = 1$, i.e. $\theta = 45^\circ$.
Maximum height: set $\dot{y} = 0$ to find $t_H = u\sin\theta / g$; substitute into $y(t)$ to get $H = \dfrac{u^2 \sin^2\theta}{2g}$.
Worked through the hook: $u = 20$, $\theta = 30^\circ$, $g = 10$.
$\ddot{x} = 0 \Rightarrow \dot{x} = u\cos\theta \Rightarrow x = u\cos\theta\cdot t$ · $\ddot{y} = -g \Rightarrow \dot{y} = u\sin\theta - gt \Rightarrow y = u\sin\theta\cdot t - \tfrac12 gt^2$ · Level-ground formulas: $T = \dfrac{2u\sin\theta}{g}$, $R = \dfrac{u^2\sin 2\theta}{g}$, $H = \dfrac{u^2 \sin^2\theta}{2g}$ · Maximum range at $\theta = 45^\circ$
Pause, copy $\ddot{x}=0$, $\ddot{y}=-g$, the parametric equations, and the level-ground formulas $T$, $R = u^2\sin 2\theta/g$, $H$, and maximum range at $\theta=45°$ into your book.
Quick check: A projectile is launched from ground level at $u = 40$ m/s, $\theta = 45^\circ$, $g = 10$. What is its range?
We just saw the projectile equations: $x = u\cos\theta\cdot t$, $y = u\sin\theta\cdot t - \tfrac{1}{2}gt^2$, giving $T = 2u\sin\theta/g$, $R = u^2\sin 2\theta/g$, $H = u^2\sin^2\theta/(2g)$.
We just saw the projectile equations: $x = u\cos\theta\cdot t$, $y = u\sin\theta\cdot t - \tfrac{1}{2}gt^2$, giving $T = 2u\sin\theta/g$, $R = u^2\sin 2\theta/g$, $H = u^2\sin^2\theta/(2g)$. That raises a question: what Cartesian curve does the projectile trace, and what does its shape tell us? This card answers it → eliminating $t$ gives $y = x\tan\theta - gx^2/(2u^2\cos^2\theta)$, a downward parabola, symmetric about $x = R/2$, with slope $\tan\theta$ at launch.
From $x = u\cos\theta\cdot t$, solve $t = \dfrac{x}{u\cos\theta}$ and substitute into $y(t)$:
This is a downward-opening parabola through the origin. Useful properties:
Trajectory: $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$, a parabola · Apex at $x = R/2$ on level ground; symmetric trajectory · At apex: $\dot{y} = 0$, $\dot{x} = u\cos\theta$, velocity is horizontal · Slope at launch is $\tan\theta$ as expected
Pause, copy the trajectory equation $y = x\tan\theta - gx^2/(2u^2\cos^2\theta)$, the parabola symmetry about $x = R/2$, the apex conditions ($\dot{y}=0$, velocity horizontal), and the launch slope $\tan\theta$ into your book.
Did you get this? True or false: at the highest point of a projectile trajectory (no air resistance), the speed of the particle is zero.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
A particle is projected from level ground with $u = 30$ m/s at $\theta = 60^\circ$. Take $g = 10$. Find the time of flight, range, and maximum height.
A ball is launched with $u = 25$ m/s at $\theta = 53^\circ$ above horizontal ($\sin 53^\circ \approx 0.8$, $\cos 53^\circ \approx 0.6$). Take $g = 10$. Find the velocity vector at $t = 3$ s.
A particle is launched from the origin with $u = 20$ m/s at $\theta = 45^\circ$. Take $g = 10$. (a) Derive the Cartesian trajectory $y(x)$. (b) Does the trajectory pass through the point $(20, 5)$?
Fill the gap: Range on level ground is $R =$ . Maximum height is $H =$ . Maximum range is achieved when $\theta =$ °.
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: on level ground, the maximum range for a given launch speed $u$ is achieved at launch angle $\theta = 45^\circ$.
Activities · practice with the ideas
Activities · practice with the ideas
A projectile is launched at $u = 50$ m/s, $\theta = 30^\circ$, $g = 10$. Find time of flight, range, and maximum height.
A ball thrown horizontally from a 20 m cliff with speed 15 m/s ($g = 10$). When does it land, and how far from the base?
Show that the trajectory of a projectile launched from the origin with speed $u$ at angle $\theta$ is $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$.
A projectile has $u = 40$ m/s, $\theta = 45^\circ$, $g = 10$. Find the velocity (magnitude and direction) at $t = 2$ s.
Show that two launch angles, $\theta$ and $90^\circ - \theta$, give the same range on level ground. Explain physically why this is so.
Odd one out: Three of these statements about projectile motion (no air resistance) are correct. Which one is NOT?
Earlier you predicted the time of flight, range and max height for a $u = 20$ m/s, $\theta = 30^\circ$ launch.
The equations gave $T = 2$ s, $R = 20\sqrt 3 \approx 34.6$ m, $H = 5$ m, all proportional to $u^2$ when $\theta$ is fixed. The big takeaway: the two axes are independent, and coupling happens only through the shared time variable. That decoupling is what makes the formulas neat. When we add air resistance next lesson, $\ddot{x}$ is no longer zero, but the philosophy of resolving and integrating axis-by-axis is identical.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A particle is projected from level ground with speed $u = 20$ m/s at $\theta = 30^\circ$. Take $g = 10$. Find the time of flight and the range. (2 marks)
Q2. A stone is thrown from the top of a 25 m cliff with $u = 20$ m/s at $\theta = 45^\circ$ above horizontal. Taking $g = 10$, find the time taken to hit the ground at the base of the cliff. (3 marks)
Q3. A projectile is fired from the origin with speed $u$ at angle $\theta$ above horizontal. (a) Derive the Cartesian trajectory equation $y(x)$. (b) Show that on level ground the maximum range is $u^2/g$, attained at $\theta = 45^\circ$. (3 marks)
Activity answers:
1. $T = \dfrac{2 \times 50 \times 0.5}{10} = 5$ s. $R = \dfrac{2500 \times \sin 60^\circ}{10} = 125\sqrt 3 / \ldots$ giving $R = 125\sqrt 3 \approx 216.5$ m (using $\sin 60^\circ = \sqrt 3/2$, so $R = 2500 \times \tfrac{\sqrt 3}{2} / 10 = 125\sqrt 3$). $H = \dfrac{2500 \times 0.25}{20} = 31.25$ m.
2. Take down positive for the fall: $20 = \tfrac12 \times 10 \times t^2 \Rightarrow t = 2$ s. Horizontal range = $15 \times 2 = 30$ m.
3. From $x = u\cos\theta\cdot t$, $t = x/(u\cos\theta)$. Then $y = u\sin\theta \cdot \dfrac{x}{u\cos\theta} - \tfrac12 g \left(\dfrac{x}{u\cos\theta}\right)^2 = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$.
4. $u_x = u_y = 20\sqrt 2$. At $t = 2$: $\dot{x} = 20\sqrt 2$, $\dot{y} = 20\sqrt 2 - 20$. Speed $= \sqrt{(20\sqrt 2)^2 + (20\sqrt 2 - 20)^2} = \sqrt{800 + 800 - 800\sqrt 2 + 400} = \sqrt{2000 - 800\sqrt 2}$. Numerically $\dot{x} \approx 28.28$, $\dot{y} \approx 8.28$, speed $\approx 29.5$ m/s, angle $\approx \arctan(8.28/28.28) \approx 16.3^\circ$ above horizontal.
5. $R(\theta) = \dfrac{u^2 \sin 2\theta}{g}$ and $R(90^\circ - \theta) = \dfrac{u^2 \sin(180^\circ - 2\theta)}{g} = \dfrac{u^2 \sin 2\theta}{g}$. Same range. Physically: low-angle launch has high horizontal speed but short flight time; high-angle launch has lower horizontal speed but proportionally longer flight time, balancing out.
Q1 (2 marks): $T = 2u\sin\theta/g = 2 \times 20 \times 0.5 / 10 = 2$ s [1]. $R = u^2\sin 2\theta / g = 400 \times \sin 60^\circ / 10 = 20\sqrt 3 \approx 34.6$ m [1].
Q2 (3 marks): $u_y = 20\sin 45^\circ = 10\sqrt 2$ [1]. Take launch level as $y = 0$ and ground as $y = -25$: $-25 = 10\sqrt 2 \cdot t - 5t^2$, i.e. $t^2 - 2\sqrt 2 \cdot t - 5 = 0$ [1]. Quadratic formula: $t = \sqrt 2 + \sqrt 7 \approx 4.06$ s (taking the positive root) [1].
Q3 (3 marks): (a) From $x = u\cos\theta\cdot t$ and $y = u\sin\theta\cdot t - \tfrac12 gt^2$, eliminate $t$: $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ [1]. (b) Setting $y = 0$ with $x \neq 0$ gives $x = R = u^2\sin 2\theta / g$ [1]. Maximum when $\sin 2\theta = 1$, i.e. $\theta = 45^\circ$, giving $R_{\max} = u^2/g$ [1].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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