CLASSICAL-MECHANICS

Setting Up Projectile Motion

IKey Formulas

Formula Name Description
v⃗0=vx0 ı^+vy0 ȷ^\vec{v}_0 = v_{x0}\,\hat{\imath} + v_{y0}\,\hat{\jmath} Velocity decomposition Split the slanted launch velocity into perpendicular pieces
vx0=v0cos⁡θ0v_{x0} = v_0 \cos\theta_0 Horizontal component Speed along the ground
vy0=v0sin⁡θ0v_{y0} = v_0 \sin\theta_0 Vertical component Speed straight up
v0=vx02+vy02,θ0=tan⁡−1 ⁣(vy0vx0)v_0 = \sqrt{v_{x0}^2 + v_{y0}^2}, \quad \theta_0 = \tan^{-1}\!\left(\dfrac{v_{y0}}{v_{x0}}\right) Magnitude & angle Recover speed and launch angle from components
F⃗g=−mg ȷ^\vec{F}_g = -mg\,\hat{\jmath} Gravitational force The entire free-body diagram (no horizontal force)

IIInteractive Visualizations

Visualization 1 — Resolving v⃗0\vec{v}_0 into Components

How do the legs vx0=v0cos⁡θ0v_{x0} = v_0\cos\theta_0 and vy0=v0sin⁡θ0v_{y0} = v_0\sin\theta_0 change as you tilt the launch velocity?

Visualization 2 — The Ideal Trajectory

With only gravity acting, what path does the projectile trace, and how does the launch angle reshape its range?

💡 On level ground the range peaks at a 45° launch; raising y0y_0 pushes the range-maximizing angle below 45°.

Visualization 3 — The Free-Body Diagram

Doubling the mass doubles the force F⃗g\vec{F}_g — so why does every projectile fall with the same acceleration?

💡 Because the only force is vertical, vxv_x stays constant while just vyv_y changes — the two axes decouple and can be solved separately.

IIIQuiz Questions

Question 1

A projectile is launched with initial speed v0=20 m/sv_0 = 20\ \text{m/s} at an angle θ0=30°\theta_0 = 30° above the horizontal.

What is the horizontal component of the initial velocity, vx0v_{x0}?

(Use cos⁡30°≈0.866\cos 30° \approx 0.866, sin⁡30°=0.5\sin 30° = 0.5.)

✅ Correct! vₓ₀ = v₀ cos θ₀ = 20 · 0.866 ≈ 17.3 m/s.

❌ Not quite. The horizontal component uses cosine: vₓ₀ = v₀ cos θ₀. (Sine gives the vertical component.)

Show solution

Solution:

The horizontal component uses the cosine:

vx0=v0cos⁡θ0=20×cos⁡30°=20×0.866≈17.3 m/sv_{x0} = v_0 \cos\theta_0 = 20 \times \cos 30° = 20 \times 0.866 \approx 17.3\ \text{m/s}

Choosing vy0=10v_{y0} = 10 would mean you used sin⁡\sin (that is the vertical component).

Question 2

A student is given v0=40 m/sv_0 = 40\ \text{m/s} and θ0=60°\theta_0 = 60° and needs the vertical component vy0v_{y0}. They write vy0=v0cos⁡θ0v_{y0} = v_0 \cos\theta_0.

Which statement is correct?

(Use cos⁡60°=0.5\cos 60° = 0.5, sin⁡60°≈0.866\sin 60° \approx 0.866.)

✅ Correct! The vertical component uses sine: vᵧ₀ = v₀ sin θ₀ = 40 · 0.866 ≈ 34.6 m/s. The student swapped sin and cos.

❌ Not quite. The vertical component uses sine: vᵧ₀ = v₀ sin θ₀. The student's cos formula gives the horizontal piece.

Show solution

Solution:

The vertical leg of the velocity triangle is opposite the angle θ0\theta_0, so it uses the sine:

vy0=v0sin⁡θ0=40×sin⁡60°=40×0.866≈34.6 m/sv_{y0} = v_0 \sin\theta_0 = 40 \times \sin 60° = 40 \times 0.866 \approx 34.6\ \text{m/s}

Using cosine (v0cos⁡θ0=20v_0\cos\theta_0 = 20) gives the horizontal component instead — a classic mix-up. The two formulas only agree at θ0=45°\theta_0 = 45°, where sin⁡=cos⁡\sin = \cos.

Question 3

A projectile has velocity components vx0=3 m/sv_{x0} = 3\ \text{m/s} and vy0=4 m/sv_{y0} = 4\ \text{m/s}.

What is the launch speed v0v_0 (the magnitude of the initial velocity)?

✅ Correct! v₀ = √(3² + 4²) = √25 = 5 m/s.

❌ Not quite. Perpendicular components combine as v₀ = √(vₓ₀² + vᵧ₀²), not by adding them directly.

Show solution

Solution:

The speed is the magnitude of the velocity vector, found with the Pythagorean theorem:

v0=vx02+vy02=32+42=9+16=25=5 m/sv_0 = \sqrt{v_{x0}^2 + v_{y0}^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\ \text{m/s}

Adding the components directly (3+4=73 + 4 = 7) is wrong — perpendicular legs combine as a square root of the sum of squares, not a plain sum. The value 2525 forgets the square root.

Question 4

In this setup we neglect air resistance and keep only gravity, choosing the yy-axis pointing up.

True or False: After launch, the projectile experiences a horizontal force that gradually slows its downrange (horizontal) motion.

✅ Correct! Gravity is the only force, Fᵍ = −mg ĵ, which is purely vertical. There is no horizontal force, so vₓ is constant.

❌ Not quite. A horizontal slowing would require air resistance, which we dropped. With gravity only (Fᵍ = −mg ĵ), there is no horizontal force and vₓ stays constant.

Show solution

Solution:

False. With air resistance dropped, the only force is gravity:

F⃗g=−mg ȷ^\vec{F}_g = -mg\,\hat{\jmath}

This points straight down and has no horizontal component. So the horizontal velocity vxv_x stays constant throughout the flight — nothing slows the downrange motion. It is precisely this absence of a horizontal force that lets the xx and yy motions decouple and be solved independently. (A horizontal slowing would require air drag, which we deliberately excluded.)

Solved: 0 / 4