Physics · Mechanics

Projectile Motion

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Wrap-up

What you learned

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Read next: Momentum and ImpulseConnect launch changes to impulse and momentum.

Key takeaway

A projectile can be modeled as horizontal motion plus vertical motion under gravity.
With drag off, horizontal velocity stays constant while vertical velocity changes steadily downward.
At the apex, vertical velocity is zero for an instant but horizontal motion continues.
Range depends on both horizontal speed and airtime, so a steeper launch is not automatically farther.

Common misconception

Do not treat the curved path as one all-purpose motion or assume the projectile stops at the top; only the vertical component reaches zero there.

Range depends on the tradeoff between horizontal speed and time in the air, not on launch angle alone.

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Equation snapshotComponent motion snapshot · Vertical positionShow 3 equations

Use x(t) for the steady sideways part and y(t) for the gravity-bent vertical part, then connect their combination to the range formula.

Component motion snapshot
$x (t) = v_{0} cos (θ) t$
With no horizontal acceleration in the ideal model, horizontal position changes linearly in time.
Vertical position
$y (t) = v_{0} sin (θ) t - \frac{1}{2} g t^{2}$
Vertical position combines the upward launch part with the downward pull of gravity.
Range
$R = \frac{v _{0}^{2} sin ( 2 θ )}{g}$
This gives the landing distance for the ideal equal-height case. At fixed speed, complementary launch angles share the same range.

Why it behaves this way

Explanation

After launch, the projectile is no longer being pushed in the ideal model. Gravity acts only downward, so the motion splits into two simpler parts: constant horizontal velocity and vertical motion with constant downward acceleration.

The stage and graphs are designed to show that same split from several angles. The trajectory is the combined result, while the component and velocity graphs show why the arc forms: the horizontal part stays steady, the vertical part slows on the way up, reaches zero at the apex, then grows downward on the way down.

Key ideas

01Treat projectile motion as two linked motions: horizontal motion with constant velocity and vertical motion with constant downward acceleration.

02At the apex, the vertical velocity is zero for an instant, but the horizontal velocity is still nonzero, so the projectile keeps moving forward.

03For equal launch and landing height, complementary launch angles give the same range, and 45° gives the maximum range for a fixed speed.

Worked examples

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Frozen walkthrough

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Frozen walkthrough

Use the current launch as evidence. First compute the range from the same speed, angle, and gravity used by the stage. Then freeze any instant in flight and read the velocity components from the same state shown on the graphs.

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Example 1 of 2

Frozen valuesUsing frozen parameters

For the current launch settings, what landing range does the equal-height formula predict?

v_{0}

Launch speed

18 m/s

θ

Launch angle

45 °

g

Gravity

9.8 m/s²

1. Use the equal-height range formula

For equal launch and landing height, use

R = \frac{v _{0}^{2} sin ( 2 θ )}{g}

2. Insert the current launch values

R = \frac{( 18 ) ^{2} sin ( 2 \cdot 4 5 ^{\circ} )}{9} .8

3. Evaluate the range

With

sin (2 θ) = 1

, the predicted range becomes

R = 33.06 m

Predicted landing range

R = 33.06 m

This angle keeps a fairly balanced split between horizontal reach and airtime, so the predicted range stays strong.

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Step 2 of 3

Motion and Circular Motion

Projectile Motion appears later in this track, so it is cleaner to start from the beginning first.

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