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Projectile Motion: The Complete Guide with Equations and Worked Examples

Dr. James CarterDr. James CarterUpdated Mar 1, 202615 min read
Projectile motion — basketball tracing a parabolic arc through the air

A basketball arcing toward the hoop, a cannonball launched from a fortress, a stream of water from a garden hose — all are examples of projectile motion. It's one of the first topics you encounter in classical mechanics, and one of the most elegant: an object moves through space under the influence of gravity alone, tracing a perfect parabolic path that's been understood mathematically since Galileo. Projectile motion is also one of the most direct applications of Newton's laws.

What Is Projectile Motion?

Projectile motion is the motion of an object launched into the air, subject only to the acceleration due to gravity (neglecting air resistance). The key insight — the idea that makes projectile motion tractable — is that horizontal and vertical motion are completely independent of each other.

Gravity acts only vertically (downward), so it affects only the vertical component of motion. The horizontal component is unaffected by gravity and therefore remains constant throughout the flight. This independence of perpendicular components is not obvious, but it is a direct consequence of Newton's second law applied component by component: the only force is gravity (vertically), so only the vertical acceleration is non-zero.

Setting Up the Equations

If an object is launched with initial speed v₀ at angle θ above the horizontal, its initial velocity components are:

v₀ₓ = v₀ cos θ      v₀ᵧ = v₀ sin θ

The horizontal and vertical positions at time t are:

x(t) = v₀ₓ · t = (v₀ cos θ) · t
y(t) = v₀ᵧ · t − ½gt² = (v₀ sin θ) · t − ½gt²

And the velocity components at time t:

vₓ(t) = v₀ cos θ   (constant — no horizontal force)
vᵧ(t) = v₀ sin θ − gt

Notice that the horizontal velocity never changes. The vertical velocity changes linearly with time at rate g = 9.8 m/s², decreasing on the way up and increasing (in the downward direction) on the way down. At the peak, vᵧ = 0 — but the object is still moving horizontally at v₀ cos θ.

The Parabolic Trajectory

If you eliminate the time variable between the x(t) and y(t) equations, you get a relationship between y and x that is quadratic — meaning the trajectory is a parabola. This is one of Galileo's great insights: the path of a projectile under uniform gravity is precisely a parabolic curve, and the shape depends only on launch angle and initial speed.

The Four Key Results

For a projectile launched from and returning to the same height:

Time to maximum height: At the top, vᵧ = 0. Setting v₀ sin θ − gt = 0:

ttop = v₀ sin θ / g

Maximum height:

H = v₀² sin²θ / (2g)

Total time of flight: By symmetry of the parabola, the descent takes as long as the ascent:

T = 2v₀ sin θ / g

Horizontal range:

R = v₀² sin(2θ) / g

This last equation reveals something elegant: for a given launch speed, the range is maximized when sin(2θ) = 1, which occurs at θ = 45°. Launch angles equally above and below 45° produce the same range — a 30° launch goes just as far as a 60° launch, though the 60° launch goes much higher and takes longer.

Energy Perspective on Projectile Motion

You can also analyze projectile motion using conservation of energy. At any point in the trajectory, the total mechanical energy is constant (no friction):

½mv² + mgy = ½mv₀²

This immediately gives you the speed at any height without tracking the time or direction of motion. At the peak (maximum height H), the kinetic energy equals the launch kinetic energy minus the potential energy gained:

vtop = v₀ cos θ

This confirms what we found kinematically: only the horizontal component remains at the peak. The energy method gives us the same answer with different (and often faster) algebra.

Common Mistakes

Mistake 1: Thinking velocity is zero at the top. Only the vertical component is zero. The horizontal speed v₀ cos θ persists throughout the entire flight, including the peak. The object is never truly stationary unless launched straight up (θ = 90°).

Mistake 2: Applying the range formula to unequal heights. The formula R = v₀² sin(2θ)/g only works when the landing height equals the launch height. For a ball thrown off a cliff, you must use the full kinematic equations with the appropriate boundary conditions.

Mistake 3: Forgetting that g is constant throughout. At the peak, gravity doesn't turn off or reduce. The acceleration is always 9.8 m/s² downward — on the way up, at the peak, and on the way down. This is a consequence of Newton's first and second laws: the net force (gravity) is constant, so the acceleration is constant.

Worked Example

A ball is launched at 20 m/s at 35° above horizontal. Find its maximum height and range.

v₀ₓ = 20 cos 35° ≈ 16.4 m/s | v₀ᵧ = 20 sin 35° ≈ 11.5 m/s

H = (11.5)² / (2 × 9.8) ≈ 6.7 m

R = (20)² × sin(70°) / 9.8 ≈ 38.4 m

Real-World Applications

Projectile motion is more than an academic exercise. It's the physics behind ballistics, sports science, civil engineering (the parabolic shape of suspension bridge cables), aerospace (initial rocket trajectory phases), water fountain design, and forensic science (reconstructing trajectories from impact sites). The principles scale from a table-tennis ball to an intercontinental ballistic missile — the physics is the same, governed throughout by F = ma applied to a constant gravitational force.

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Dr. James Carter

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Dr. James Carter

Physicist and educator with 15+ years teaching classical mechanics and thermodynamics at the university level. Former MIT OpenCourseWare contributor.

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