Projectile Motion: Launch at an Angle

Fire the cannon and see why the body flies along a parabola. Its motion is made of two simple ones: horizontally it moves at a constant speed, while vertically it slows down and speeds up like a ball thrown straight up. Shadows on the axes and a strobe make this visible, and the numbers on the display match the textbook formulas.

Motion graphs

The dashed line is the whole flight from the formulas, the solid line is drawn together with the body. Compare: vₓ stays the same, while vᵧ decreases steadily and passes through zero at the top.

v₀ initial speed 15 m/s

θ launch angle to horizontal 50°

h₀ launch height 0 m

Where we throw

g free-fall acceleration 9,8 m/s²

Ready-made experiments

Playback speed

Show

where it will fly (dashed)

velocity v and acceleration g vectors

velocity components vₓ and vᵧ

strobe: positions at equal Δt

shadows of the body on the x and y axes

compare with a ball in air

Game: hit the target

Air resistance is ignored (except for the orange ball); the body is launched from height h₀ above flat ground. Position and velocity come from the exact formulas, so the numbers on the screen match the textbook.

The key idea in short: a launched body takes part in two motions at once. Horizontally it moves uniformly — vₓ = v₀·cos θ does not change. Vertically it moves with constant acceleration g, like a ball tossed straight up. Together these two motions make a parabola. Example: v₀ = 20 m/s, θ = 30° — the body rises 5.1 m and lands after 2.04 s, 35.4 m from the cannon. Set these numbers and check.

KEY FORMULAS

Components of the initial velocity

$$v_{0x} = v_0\cos\theta, \qquad v_{0y} = v_0\sin\theta$$

The initial velocity is split into horizontal and vertical parts. θ is the angle between the velocity vector and the horizontal.

▶ Detailed Explanation

The vector v₀ is the hypotenuse of a right triangle, and its components are the legs. The horizontal leg is adjacent to the angle θ, so v₀ₓ = v₀·cos θ; the vertical leg is opposite the angle, so v₀ᵧ = v₀·sin θ. Check: v₀ₓ² + v₀ᵧ² = v₀².

Example: v₀ = 20 m/s, θ = 30°. v₀ₓ = 20·0.866 = 17.32 m/s, v₀ᵧ = 20·0.5 = 10 m/s. Before the launch the simulator draws these two arrows right at the cannon.

Horizontally — uniform motion

$$v_x = v_0\cos\theta = \text{const}, \qquad x = v_0\cos\theta \cdot t$$

No force acts on the body horizontally, so the horizontal velocity does not change during the whole flight.

▶ Detailed Explanation

Gravity points vertically, its horizontal component is zero, so aₓ = 0. Every second the body moves forward by the same distance vₓ. In the simulator you can see this in the blue shadow on the ground: the strobe marks on the x-axis are equally spaced. On the vₓ(t) graph it is a horizontal line.

Vertically — uniformly accelerated motion

$$v_y = v_0\sin\theta - g t, \qquad y = h_0 + v_0\sin\theta \cdot t - \frac{g t^2}{2}$$

The vertical velocity decreases by g every second: vertically the body moves just like one thrown straight up. h₀ is the launch height.

▶ Detailed Explanation

The y-axis points up and the acceleration g points down, hence the minus sign. While vᵧ > 0 the body rises and slows down vertically; at the top vᵧ = 0; then vᵧ < 0 and the body falls faster and faster. The vᵧ(t) graph is a sloped line with slope −g, the y(t) graph is a parabola.

Example: v₀ᵧ = 10 m/s, h₀ = 0. After 0.5 s vᵧ = 10 − 9.8·0.5 = 5.1 m/s, y = 10·0.5 − 9.8·0.25/2 = 3.78 m.

Rise time and maximum height

$$t_1 = \frac{v_0\sin\theta}{g}, \qquad H = h_0 + \frac{v_0^2\sin^2\theta}{2g}$$

At the top the vertical velocity is zero. The height depends only on the vertical component of the initial velocity.

▶ Detailed Explanation

From vᵧ = v₀·sin θ − g·t₁ = 0 we get t₁. Substituting t₁ into y(t) gives H. At the top the speed of the body is not zero but vₓ = v₀·cos θ, and the acceleration is still g.

Example: v₀ = 20 m/s, θ = 30°, h₀ = 0. t₁ = 10/9.8 = 1.02 s, H = 10²/(2·9.8) = 5.10 m. At θ = 90° (straight up) H = v₀²/(2g) = 20.4 m — the limit for this speed.

Flight time (launch from the ground)

$$T = \frac{2 v_0 \sin\theta}{g}$$

If the body is launched from the ground and lands at the same level, the rise and the fall take the same time, so T = 2t₁.

▶ Detailed Explanation

The formula follows from y(T) = v₀·sin θ·T − g·T²/2 = 0. The flight time depends only on the vertical component of the velocity: bodies launched with the same v₀ᵧ stay in the air equally long for any vₓ.

Example: v₀ = 20 m/s, θ = 30°: T = 2·10/9.8 = 2.04 s. If the body is launched from a height h₀ this formula does not apply — see the formula for a launch from a height.

Range (launch from the ground)

$$L = v_{0x}\,T = \frac{v_0^2 \sin 2\theta}{g}, \qquad L_{max} = \frac{v_0^2}{g}\ (\theta = 45^\circ)$$

The range is the horizontal speed times the flight time. It is largest at 45°; the angles θ and 90° − θ give the same range.

▶ Detailed Explanation

L = v₀·cos θ · 2v₀·sin θ / g = v₀²·2sin θ·cos θ / g = v₀²·sin 2θ / g. A sine is never more than one, and sin 2θ = 1 at θ = 45°. Since sin 2θ = sin(180° − 2θ), the angles 30° and 60° give the same range.

Example: v₀ = 20 m/s, θ = 30°: L = 400·0.866/9.8 = 35.35 m; at 45° L_max = 400/9.8 = 40.8 m. The L(θ) graph under the scene shows the whole curve.

Equation of the path — a parabola

$$y = h_0 + x\tan\theta - \frac{g\,x^2}{2 v_0^2 \cos^2\theta}$$

Expressing the time from x(t) and substituting it into y(t) gives y as a quadratic function of x — a downward-opening parabola.

▶ Detailed Explanation

t = x / (v₀·cos θ). Substituting into y = h₀ + v₀·sin θ·t − g·t²/2 gives the formula. The first term with x is the straight line the body would follow without gravity (along the cannon barrel). The second is how far the body “drops below” that line because of the fall: exactly g·t²/2.

The equation helps in problems like “will the body clear a wall of height h at distance x”: substitute x and compare y with h.

Launch from a height h₀

$$T = \frac{v_0\sin\theta + \sqrt{v_0^2\sin^2\theta + 2 g h_0}}{g}, \qquad L = v_0\cos\theta \cdot T$$

The flight time is the positive root of y(T) = 0. In a horizontal launch (θ = 0) T = √(2h₀/g), the same as for a body simply dropped.

▶ Detailed Explanation

The equation h₀ + v₀ᵧ·T − g·T²/2 = 0 is quadratic in T. The negative root is discarded. With h₀ = 0 the formula becomes T = 2v₀·sin θ / g.

Example: from a tower h₀ = 20 m a body is thrown horizontally at 15 m/s. T = √(2·20/9.8) = 2.02 s, L = 15·2.02 = 30.3 m. The speed at the ground is v = √(v₀² + 2gh₀) = √(225 + 392) = 24.8 m/s. From a height the best angle is below 45°: tan θ* = v₀ / √(v₀² + 2gh₀).

Speed at any point of the path

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 - 2g\,(y - h_0)}, \qquad \tan\alpha = \frac{v_y}{v_x}$$

The speed depends only on the height above the launch point. α is the angle the velocity vector makes with the horizontal at that moment.

▶ Detailed Explanation

The second form is the law of conservation of energy: mv₀²/2 + mgh₀ = mv²/2 + mgy. So at the same height on the way up and on the way down the speed is the same, and with h₀ = 0 the body lands at speed v₀. The smallest speed is at the top and equals vₓ = v₀·cos θ.

Example: v₀ = 20 m/s, θ = 30°: at the top v = 17.32 m/s, at the ground it is 20 m/s again, at 30° downward.

What projectile motion is. A body is launched with speed v₀ at an angle θ to the horizontal, and after that only gravity acts on it. Air resistance is neglected. Then the acceleration of the body is always the same: a = g ≈ 9.8 m/s², pointing straight down. This is true on the way up, at the top and on the way down. A stone, a ball, a jet of water from a hose and a cannonball all move like this. The path of such a motion is a parabola.

Two motions instead of one. The key idea, noticed by Galileo: horizontal and vertical motion do not interfere with each other. The initial velocity is split into two components: v₀ₓ = v₀·cos θ and v₀ᵧ = v₀·sin θ. No force acts horizontally, so vₓ = v₀ₓ stays constant and x = v₀ₓ·t. This is uniform motion. Gravity acts vertically, so vᵧ = v₀ᵧ − g·t and y = h₀ + v₀ᵧ·t − g·t²/2. This is uniformly accelerated motion, exactly like that of a body thrown straight up at speed v₀ᵧ. In the simulator you can see it in the shadows: the blue shadow on the ground covers equal distances in equal Δt, while the green shadow on the y-axis rises more and more slowly, stops and falls faster and faster.

Velocity during the flight. The velocity vector always points along the tangent to the path. Its horizontal part vₓ does not change, while the vertical part vᵧ decreases by 9.8 m/s every second. At the top vᵧ = 0, but the velocity itself is not zero: the body keeps flying forward at speed vₓ. This is the smallest speed of the whole flight. With h₀ = 0 the body hits the ground with the same speed v₀ it was launched with and at the same angle θ, only pointing down. The speed at any point is v = √(vₓ² + vᵧ²) = √(v₀² − 2g(y − h₀)). It depends only on the height, not on whether the body is going up or down.

Height, time and range. The rise time follows from vᵧ = 0: t₁ = v₀·sin θ / g. The maximum height is H = v₀²·sin²θ / (2g). For a launch from the ground the rise and the fall take the same time, so the total flight time is T = 2v₀·sin θ / g. The range is the horizontal speed times the flight time: L = v₀ₓ·T = v₀²·sin 2θ / g. Example: v₀ = 20 m/s, θ = 30°. Then v₀ₓ = 17.32 m/s, v₀ᵧ = 10 m/s, t₁ = 1.02 s, H = 5.10 m, T = 2.04 s, L = 35.35 m. Set these values in the simulator and check every number in the Measurements panel.

Why 45°. The range formula contains sin 2θ, which is largest when 2θ = 90°, that is θ = 45°. Then L_max = v₀²/g. Angles that add up to 90° (30° and 60°, 15° and 75°) give the same range, because sin 60° = sin 120°. But at the larger angle the body flies higher and longer. You can see it on the right-hand graph below the scene: the curve L(θ) is symmetric about 45°. If you launch from a height h₀, the best angle becomes smaller than 45°: from a 10 m tower at v₀ = 20 m/s the best angle is 39.3°, and the body flies 49.8 m instead of 40.8 m from the ground.

Launch from a height and horizontal launch. If the body is launched from a height h₀, the flight time follows from the equation y(T) = 0: T = (v₀ᵧ + √(v₀ᵧ² + 2gh₀))/g, and the range is still L = v₀ₓ·T. The formula L = v₀²·sin 2θ / g no longer works here. In a horizontal launch (θ = 0) there is no vertical velocity at the start, and the body falls for exactly as long as it would if it were simply dropped: T = √(2h₀/g). From 20 m that is 2.02 s for any horizontal speed. Check it with the “Throw vs drop” experiment.

Other planets and air. Every formula has g in the denominator. On the Moon g = 1.62 m/s², six times less than on Earth, so a body launched at the same speed flies six times higher and six times farther. On Jupiter (g = 24.8 m/s²) it is the other way round: 2.5 times shorter. In real air a ball does not follow a parabola: drag takes away its speed, the path becomes steeper on the way down, and the range is noticeably shorter. A football kicked at 25 m/s at 45° would fly 63.8 m in a vacuum but only about 40 m in air. Its best angle is no longer 45° but about 41–43°. Turn on “Compare with a ball in air” and look at the orange curve.

Common mistakes in problems. Thinking that the speed at the top is zero, when only vᵧ is zero. Thinking that the acceleration disappears at the top, when it is equal to g all the time. Mixing up sine and cosine: the horizontal component is v₀·cos θ, the vertical one v₀·sin θ (the angle is measured from the horizontal). Using T = 2v₀·sin θ / g and L = v₀²·sin 2θ / g for a launch from a height. Forgetting that the horizontal speed does not affect the fall time.

How the simulator works. The position and velocity of the body are calculated from the exact formulas x(t) and y(t), not approximately frame by frame, so the numbers on the display match the textbook to the last digit and do not depend on the playback speed. The moments of the top point and of landing are computed exactly. The flight of the ball in air (mass 0.43 kg, diameter 22 cm) is computed numerically with the fourth-order Runge–Kutta method. Smoke, sparks and dust are decoration only and do not affect the calculation. On Earth g is taken as 9.8 m/s².