Projectile motion · SillyChili

1Wonder

A basketball shot, a water fountain, a long jumper: they all fly along the same kind of curve. Why that shape?

Make a guess first: which launch angle sends a ball the farthest?

2Explore

Projectile lab

This activity needs a bigger screen

Turn your phone sideways, or open this lesson on a tablet or computer. The explanation and quiz below work here.

Time 0 s x 0 m y 0 m Formula predicts: R = v₀² sin 2θ / g = 40.8 m

Launch angle θ 45 ° Launch speed v₀ 20 m/s

Planet

Speed

Velocity arrows x and y shadows Air resistance

Try this

Launch at 30° and then at 60° with the same speed. Where do they land?
Find the angle that gives the longest range. Is it the same on the Moon?
Turn on x and y shadows. How does the blue shadow move? And the red one?
Turn on air resistance. Is 45° still the best angle?

3Explain

Step 1 of 5

Two motions at once

The ball moves sideways and up-and-down at the same time. Each dot is a snapshot taken every 0.25 s. Watch its two shadows: blue on the ground and red on the wall.

The equations

Split the launch speed $v_0$ into two parts using the angle $\theta$ :

\htmlClass{c-blue}{v_x = v_0\cos\theta} \qquad \htmlClass{c-red}{v_y = v_0\sin\theta - g\,t}

Then the position at time $t$ is

\htmlClass{c-blue}{x = v_0\cos\theta\; t} \qquad \htmlClass{c-red}{y = v_0\sin\theta\; t - \tfrac{1}{2} g t^2}

The ball lands when $y = 0$ again. Solve for $t$ and substitute into $x$ :

Range on flat ground (no air)

R = \frac{v_0^2 \sin 2\theta}{g}

$\sin 2\theta$ is largest (equal to 1) when $2\theta = 90^\circ$ , so 45° gives the longest range. And since $\sin(180^\circ - 2\theta) = \sin 2\theta$ , the angles $\theta$ and $90^\circ - \theta$ give the same range.

4Practice

Question 1 of 5

A ball launched at 30° lands 25 m away. Which other angle (same speed) also lands 25 m away?

5Recap

Projectile motion is two simple motions at once: steady sideways motion plus falling.
vₓ = v₀ cos θ stays constant. v_y changes by $g$ every second.
The path is a parabola.
Range: $R = \dfrac{v_0^2 \sin 2\theta}{g}$ . With the same speed, 45° goes farthest, and θ and 90° − θ land at the same place.
Air resistance makes the range shorter and the best angle a little lower than 45°.