Projectile Motion

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

📊Same 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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