Why is 45° best without air resistance, but lower with it?

From level ground with no drag the range is R = v₀² sin 2θ / g, which peaks at θ = 45°. With drag, the longer the projectile stays in the air and the farther it travels, the more speed it loses, and high lobs suffer most. So a flatter launch wins. The stronger the drag (light, large or fast balls), the lower the best angle: a ping-pong ball's can be in the low 30s or below. Launching from a height also lowers the best angle even without drag: θ* = arctan(v₀ / √(v₀² + 2gh₀)).

How is quadratic drag computed, and why numerically?

At typical ball speeds, air resistance is roughly proportional to the square of the speed, F = ½ρC_dAv², and points opposite the velocity. As an acceleration that is a = −g ĵ − k|v|v with k = ρC_dA / 2m. Because |v| contains both vₓ and v_y, the horizontal and vertical motions are coupled and there is no elementary closed form. The tool integrates with fourth-order Runge–Kutta at about 1/2000 of the flight time per step and finds the landing time (y = 0) by bisection.

What does terminal velocity mean here?

As an object falls, drag grows with speed until it balances gravity. The acceleration is then zero and the speed stops increasing: that is the terminal velocity v_t = √(g/k) = √(2mg / ρC_dA). It is the easiest way to judge how much drag matters. It's about 40 m/s for a baseball, 27 m/s for a soccer ball and only about 9 m/s for a ping-pong ball. Well below v_t drag barely changes the path. Near or above it, the trajectory becomes lopsided, with a much steeper descent.

Why is the path with drag not symmetric?

Without drag the horizontal velocity is constant, so the rise and fall mirror each other about the apex. With drag the horizontal velocity keeps falling, so after the apex the projectile covers less ground in the same time. The descent is steeper and shorter than the climb, the impact speed is lower than the launch speed and the impact angle is steeper than the launch angle. The results table shows how much energy drag took away.

What happens if I throw something on the Moon?

The Moon's gravity is about 1.62 m/s², one sixth of Earth's, and there is essentially no atmosphere. Since R = v₀² sin 2θ / g, the same throw goes about six times as far and six times as high as on Earth without drag, and stays up six times longer. Choosing the Moon preset sets the air density to zero, so you see a pure parabola.

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🎯Projectile Motion Simulator - Air Drag, Optimal Angle, Trajectories

Set the launch speed, angle and height with the sliders, or press and drag on the canvas to aim and release to fire. The projectile flies along its path while the time, position, velocity components and speed update live. Turn on air resistance to integrate quadratic drag with RK4 using realistic values for a baseball, soccer ball, ping-pong ball and more. The no-drag parabola stays on screen for comparison, and the tool searches numerically for the launch angle that gives the longest range with drag. You can keep up to 8 trajectories on the canvas and change parameters between shots to compare them.

Free · No sign-up · Runs in your browserLast updated: 2026-10-26

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How to use Projectile Motion Simulator (with Air Resistance)

1Set the launch speed, angle and height, and pick a planet (or enter your own g)
2Optionally turn on air resistance and choose a ball, or enter the mass, C_d, diameter and air density
3Press Launch, or drag on the canvas to aim and release to fire, and watch the path and velocity vectors
4Compare range, max height, flight time, impact speed and best angle with and without drag, and keep trajectories to overlay them

FAQ

Why is 45° best without air resistance, but lower with it?: From level ground with no drag the range is R = v₀² sin 2θ / g, which peaks at θ = 45°. With drag, the longer the projectile stays in the air and the farther it travels, the more speed it loses, and high lobs suffer most. So a flatter launch wins. The stronger the drag (light, large or fast balls), the lower the best angle: a ping-pong ball's can be in the low 30s or below. Launching from a height also lowers the best angle even without drag: θ* = arctan(v₀ / √(v₀² + 2gh₀)).
How is quadratic drag computed, and why numerically?: At typical ball speeds, air resistance is roughly proportional to the square of the speed, F = ½ρC_dAv², and points opposite the velocity. As an acceleration that is a = −g ĵ − k|v|v with k = ρC_dA / 2m. Because |v| contains both vₓ and v_y, the horizontal and vertical motions are coupled and there is no elementary closed form. The tool integrates with fourth-order Runge–Kutta at about 1/2000 of the flight time per step and finds the landing time (y = 0) by bisection.
What does terminal velocity mean here?: As an object falls, drag grows with speed until it balances gravity. The acceleration is then zero and the speed stops increasing: that is the terminal velocity v_t = √(g/k) = √(2mg / ρC_dA). It is the easiest way to judge how much drag matters. It's about 40 m/s for a baseball, 27 m/s for a soccer ball and only about 9 m/s for a ping-pong ball. Well below v_t drag barely changes the path. Near or above it, the trajectory becomes lopsided, with a much steeper descent.
Why is the path with drag not symmetric?: Without drag the horizontal velocity is constant, so the rise and fall mirror each other about the apex. With drag the horizontal velocity keeps falling, so after the apex the projectile covers less ground in the same time. The descent is steeper and shorter than the climb, the impact speed is lower than the launch speed and the impact angle is steeper than the launch angle. The results table shows how much energy drag took away.
What happens if I throw something on the Moon?: The Moon's gravity is about 1.62 m/s², one sixth of Earth's, and there is essentially no atmosphere. Since R = v₀² sin 2θ / g, the same throw goes about six times as far and six times as high as on Earth without drag, and stays up six times longer. Choosing the Moon preset sets the air density to zero, so you see a pure parabola.