At what angle is range maximum (no air resistance)?

R = v₀² sin(2θ)/g is maximum when sin(2θ)=1 → θ=45°.

From x-t data, determine the launch speed if angle=30°.

V_x = v₀cos(30°); slope of x vs t gives v_x; v₀ = v_x/cos(30°).

With air resistance, why does the optimal angle drop below 45°?

Horizontal deceleration means spending less time in air is better; lower angle = faster horizontal motion.

Pro 🔒~25 min

Projectile Data Lab

Collect and analyze real projectile motion data

How it works

Projectile motion separates into independent horizontal (constant velocity) and vertical (constant acceleration −g) components. Without air resistance, the optimal range angle is 45°. Air resistance reduces range and optimal angle (below 45°). By collecting position-time data, you can fit parabolic curves and extract launch speed and angle even without knowing them in advance.

Upgrade to Pro to access this experiment

Step-by-step

Set launch parameters and fire the projectile.
Click on the trajectory to sample data points.
The data table records (x, t) and (y, t) values.
Fit a parabola to the y-t data to extract g and initial vertical speed.
Enable air resistance to observe the asymmetric trajectory.

Key formulas

$x = v_0\cos\theta \cdot t$ Horizontal position
$y = v_0\sin\theta \cdot t - \frac{1}{2}gt^2$ Vertical position
$R = \frac{v_0^2\sin 2\theta}{g}$ Range (flat ground, no air resistance)

Frequently asked questions

At what angle is range maximum (no air resistance)?: R = v₀² sin(2θ)/g is maximum when sin(2θ)=1 → θ=45°.
From x-t data, determine the launch speed if angle=30°.: V_x = v₀cos(30°); slope of x vs t gives v_x; v₀ = v_x/cos(30°).
With air resistance, why does the optimal angle drop below 45°?: Horizontal deceleration means spending less time in air is better; lower angle = faster horizontal motion.