Projectile Motion

A projectile moves horizontally at constant velocity and vertically with uniform acceleration due to gravity at the same time. The two motions are independent and share only time; range is greatest at 45° and angles θ and 90°−θ match. Even at the peak, horizontal velocity remains so speed is not zero. Slide launch angle and initial speed here to watch the path, range, and maximum height.

Horizontal and vertical do not interfere

🎯 Independence of motion

①A projectile moves horizontally and vertically at the same time

②Horizontal: no force, so velocity is constant (uniform)

③Vertical: only gravity acts, so it is uniformly accelerated

④The two motions are independent and share only the time t

Change the launch angle and speed

Launch angle θ45°

Initial speed v₀20 m/s

📐 Why is 45° the farthest?

①Range R = v₀²sin2θ/g — largest when sin2θ is maximal (=1), at θ=45°

②θ and 90°−θ give the same range (e.g. 30° and 60°)

③A larger angle flies higher and longer

Horizontal and vertical motion in equations

Horizontal — uniform

x = v₀ cosθ · t, v_x = v₀ cosθ

horizontal velocity never changes (no force)

Vertical — uniformly accelerated

y = v₀ sinθ · t − 12gt², v_y = v₀ sinθ − gt

gravity g changes the vertical velocity each moment

Range, max height, time of flight

Horizontal range

R = v₀² sin2θg

maximal at θ=45°

Maximum height

H = (v₀ sinθ)²2g

at the apex v_y = 0

Time of flight

T = 2 v₀ sinθg

rise time equals fall time

Angle comparison

ChartSame v₀, by launch angle (g=10)

Angle	Range	Max height
30°	medium (sin60°)	low
45°	maximum (sin90°=1)	medium
60°	medium (sin120°=sin60°)	high

From concept to problem

Example 1 — range

An object is thrown at 20 m/s and 45°. Find its horizontal range. (g=10 m/s²)

Substitute into the range formula.

R = v₀² sin2θg

Use sin(2·45°)=sin90°=1.

R = 20² × 110 = 40

▸ 40 m

At 45°, sin2θ reaches its maximum of 1, giving the longest range.

Example 2 — speed at the apex

An object thrown at 20 m/s and 60° reaches its highest point. Find its speed and direction of motion there.

At the apex the vertical velocity v_y = 0, so only the horizontal component remains.

Find the horizontal velocity.

v_x = v₀ cosθ = 20 × cos60° = 20 × 0.5 = 10

▸ speed 10 m/s, horizontal direction

The horizontal velocity remains even at the apex, so the speed is not zero — a common mistake.

Exam Points

CSAT-style (adapted, Physics II)

With the same initial speed, object A is thrown at 30° and object B at 60°. Choose the correct statement (ignore air resistance).

①A has a larger range than B.

②A and B have the same range.

③A rises higher than B.

④A has a longer time of flight than B.

⑤At the apex the speed of both objects is zero.

▸ ② A and B have the same range.

Range R ∝ sin2θ; sin(2·30°)=sin60° and sin(2·60°)=sin120°=sin60°, so they are equal → ②.

Max height H ∝ sin²θ, so B (60°) rises higher and stays up longer → ③, ④ wrong.

At the apex the horizontal velocity v₀cosθ remains, so speed is not zero → ⑤ wrong.

🎯 Exam Points

①Horizontal = uniform (v₀cosθ), vertical = accelerated (gravity) — independence of motion

②Range R=v₀²sin2θ/g, max at 45°, θ and 90°−θ are equal

③Max height H=(v₀sinθ)²/2g, v_y=0 at apex

④Time of flight T=2v₀sinθ/g

⑤Apex speed is not zero but the horizontal part v₀cosθ

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Gravitation and Kepler's Laws

Torque and Rotational Equilibrium

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