Motion · 02 / 03

Projectile Motion

Use one shared clock to uncover the two motions hidden inside a curved trajectory.

Phenomenon

A thrown object follows a curve, but neither direction has to follow a curved rule.

Scrub through one flight below. Watch the orange horizontal velocity arrow and the violet vertical velocity arrow separately. Which one changes, and what happens to the other at the very top of the path?

Trajectory sandbox

Scrub the same clock through x and y.

t = 0.85 s
0 m20.0 m
horizontal velocity
9.9 m/s
vertical velocity
1.6 m/s
horizontal position
8.4 m
height
4.9 m
Conceptual bridge

The curve is what two independent motions look like when they share one clock.

In this ideal model there is no horizontal acceleration, so the orange velocity component stays fixed. Gravity acts vertically, so the violet component steadily decreases, reaches zero for an instant, then points downward.

The projectile does not switch from “horizontal motion” to “vertical motion.” It performs both at every moment. Their shared time coordinate is what recombines them into one trajectory.

Launch decomposition
v0x=v0cosθv_{0x}=v_0\cos\theta
v0y=v0sinθv_{0y}=v_0\sin\theta

These are the two starting velocities you were already watching in the sandbox.

What happens side to side?

Horizontal story

With no horizontal force in the ideal model, horizontal acceleration is zero and horizontal velocity stays constant.

x(t)=v0xtx(t)=v_{0x}t
vx(t)=v0xv_x(t)=v_{0x}
What happens up and down?

Vertical story

Gravity supplies constant downward acceleration, so vertical velocity changes linearly while vertical position changes quadratically.

y(t)=v0yt12gt2y(t)=v_{0y}t-\frac{1}{2}gt^2
vy(t)=v0ygtv_y(t)=v_{0y}-gt
Physics ↔ Algebra

Eliminate time and the trajectory itself becomes quadratic.

y=xtanθgx22v02cos2θy=x\tan\theta-\frac{gx^2}{2v_0^2\cos^2\theta}

The parabola follows directly from constant horizontal velocity plus constant vertical acceleration.

Common pitfall

The projectile does not stop at the top.

Only the vertical velocity is zero there. Horizontal velocity remains nonzero, and gravity is still accelerating the projectile downward.

Application · find the apex

Fresh launch: 17 m/s at 38°. Scrub until the projectile reaches its greatest height.

Use the height readout rather than memorizing a time fraction. Lock the moment you think is the apex, then inspect both velocity components.