Projectile Motion (Trajectory Parabola)
Projectile motion splits the movement into a uniform horizontal and an accelerated vertical component; together they form the trajectory parabola.
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Formula
x = v_0 \cos\alpha \cdot t \qquad y = v_0 \sin\alpha \cdot t - \frac{1}{2} g t^2Variables & units – Projectile Motion (Trajectory Parabola)
| Symbol | Meaning | Unit |
|---|---|---|
| x | Horizontal position at time t | m |
| y | Height above the launch point | m |
| v₀ | Launch speed | m/s |
| α | Launch angle to the horizontal | ° |
| t | Time since launch | s |
| g | Gravitational acceleration (9.81 m/s²) | m/s² |
Derivation & background – Projectile Motion (Trajectory Parabola)
Galileo recognised the superposition principle: horizontal and vertical motion proceed independently. No force acts horizontally (uniform motion), only gravity acts vertically (free fall). Eliminating t yields the trajectory equation y(x) = x·tanα − g·x²/(2v₀²cos²α), a downward-opening parabola. Without air resistance the range W = v₀²·sin(2α)/g is maximal at 45°, and the rise height is H = v₀²·sin²α/(2g).
Exam blueprint
Validity range
Applies without air resistance and with constant gravitational acceleration. The range and height formulas assume equal launch and landing height; for an elevated launch point, find the flight time from y(t) = 0.
Derivation steps
The motion is the superposition of two independent component motions (superposition principle).
- 1No force acts horizontally: uniform motion x = v₀·cosα·t.
- 2Vertically only gravity acts: y = v₀·sinα·t − ½gt², free fall with an initial velocity.
Rearrangements
Range
Maximal at α = 45°, where sin(2α) reaches 1.
Maximum height
At the highest point the vertical velocity is zero.
Time of flight
Holds for equal launch and landing height (twice the rise time).
Task variant
A ball is thrown at v₀ = 15 m/s and 30°. Find the range.
W = v₀²·sin(2α)/g = 225 × sin(60°)/9.81 = 225 × 0.866/9.81 ≈ 19.9 m.
How high does a ball rise at v₀ = 20 m/s and α = 60°?
H = v₀²·sin²α/(2g) = 400 × 0.75/19.62 ≈ 15.3 m, since sin²(60°) = 0.75.
Common mistakes
Using the full v₀ in one direction instead of splitting it into components.
Always decompose: v₀·cosα horizontally, v₀·sinα vertically.
Calculator set to radians while entering the angle in degrees.
Check the angle mode before calculating (DEG for degree values).
Applying the 45° rule for maximum range even with an elevated launch point.
With unequal launch and landing heights the optimal angle is below 45°.
Assuming a horizontal acceleration.
Without air resistance the horizontal velocity stays constant.
Exam context
- Typical tasks: computing range and maximum height, finding the flight time from the y equation or the landing point for an elevated launch, often combined with energy conservation.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Projectile kinematics
Builds directly on free fall and uniformly accelerated motion.
Worked example
Ball thrown at v₀ = 20 m/s and α = 45°: range W = v₀²·sin(2α)/g = 400 × 1/9.81 ≈ 40.8 m, maximum height H = v₀²·sin²α/(2g) = 400 × 0.5/19.62 ≈ 10.2 m.
Applications
Ball sports (launch angles), long jump and shot put, water fountains, ballistics, irrigation technology
Quanta exam set
Curated exam set for "Projectile Motion (Trajectory Parabola)":
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Which formula describes Projectile Motion (Trajectory Parabola)?
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How do you rearrange x = v₀·cosα·t; y = v₀·sinα·t − ½gt² for Range?
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Which common mistake happens with Projectile Motion (Trajectory Parabola)?
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Frequently asked questions about Projectile Motion (Trajectory Parabola)
How do you calculate the range of projectile motion?+
The fastest way is the range formula W = v₀²·sin(2α)/g, which holds for equal launch and landing heights. Insert the launch speed in m/s and the angle in degrees. Example: v₀ = 20 m/s, α = 45° gives W = 400 × sin(90°)/9.81 ≈ 40.8 m. Alternatively, work in two steps: first find the flight time from the vertical motion (t = 2·v₀·sinα/g), then substitute into the horizontal equation x = v₀·cosα·t. This route still works when launch and landing heights differ and the ready-made range formula fails.
Why is 45 degrees the optimal launch angle?+
In the range formula W = v₀²·sin(2α)/g the factor sin(2α) appears. The sine reaches its maximum of 1 exactly at 2α = 90°, i.e. at α = 45°. Flatter throws have plenty of horizontal speed but too little flight time; steeper throws fly long but barely move forward. 45° is the best compromise. Important: this only holds without air resistance and with equal launch and landing heights. When throwing from a height, as in shot put from shoulder level, the optimal angle is lower, typically 40° to 43°. Air resistance lowers it as well, for a football clearly below 45°.
How do you find the maximum height in projectile motion?+
At the highest point the vertical velocity is zero while the horizontal one continues unchanged. From v_y = v₀·sinα − g·t = 0 the rise time is t_s = v₀·sinα/g. Substituting into the y equation gives the closed formula H = v₀²·sin²α/(2g). Example: v₀ = 20 m/s, α = 60°: H = 400 × 0.75/19.62 ≈ 15.3 m, since sin²(60°) = 0.75. Make sure to form sin²α, first the sine, then the square. A common mistake is typing sin(α²). Alternatively, energy conservation also works: ½·v_y0² = g·H with the vertical launch speed v_y0.
What is the difference between angled and horizontal projectile motion?+
The horizontal launch is the special case α = 0: the body starts without vertical initial velocity, usually from a height h, and the equations simplify to x = v₀·t and y = h − ½gt². The fall time then depends only on the height (t = √(2h/g)), exactly as in free fall, and the range is x = v₀·√(2h/g). In angled projectile motion the vertical launch component v₀·sinα is added, so the body first rises and then falls. Both cases follow the same principle: horizontal and vertical motions run independently and are calculated separately.
Why may the two motions be treated separately?+
This is the superposition principle of kinematics: forces and accelerations act component-wise. Gravity points exactly downward and has no horizontal component. Therefore the horizontal velocity v₀·cosα stays constant during the whole flight, while the vertical motion is free fall with initial velocity v₀·sinα. Galileo confirmed this experimentally: a dropped ball and a horizontally launched ball hit the ground at the same time. The link between the two component motions lies solely in the shared time t. Eliminating t produces the parabola y(x), hence the name projectile parabola.
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Create a curated FSRS exam set for x = v₀·cosα·t; y = v₀·sinα·t − ½gt²: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
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How do you calculate with Projectile Motion (Trajectory Parabola)?
Here is how to work through a typical Projectile Motion (Trajectory Parabola) (x = v₀·cosα·t; y = v₀·sinα·t − ½gt²) task step by step:
- 1
Task
A ball is thrown at v₀ = 15 m/s and 30°. Find the range.
Solution path
W = v₀²·sin(2α)/g = 225 × sin(60°)/9.81 = 225 × 0.866/9.81 ≈ 19.9 m.
- 2
Task
How high does a ball rise at v₀ = 20 m/s and α = 60°?
Solution path
H = v₀²·sin²α/(2g) = 400 × 0.75/19.62 ≈ 15.3 m, since sin²(60°) = 0.75.