§ 4. Projectile Launched at an Angle to the Horizontal (Grade 9–10)
Vacuum parabola vs. ballistic trajectory with air drag. Ballistic motion in a uniform gravitational field is a superposition of uniform horizontal motion along Ox and uniformly accelerated vertical motion along Oy.
Theory Summary & Scientific Mechanism
Projectile motion launched at angle α with initial speed v₀ illustrates Galileo’s principle of independence of motions. In vacuum (k = 0), no horizontal force acts along Ox, keeping vₓ = v₀cos α = const, while along the vertical axis Oy gravity produces constant downward acceleration a_y = -g.
Eliminating time t = (x)/(v₀cos α) from x(t) = v₀cos α · t and y(t) = v₀sin α · t - (1)/(2)gt² yields the parabolic trajectory equation: y(x) = xtan α - (gx²)/(2v₀²cos ²α). The time of flight is t_flight = (2v₀sin α)/(g), the apex height is H = (v₀²sin ²α)/(2g), and the horizontal range L = (v₀²sin (2α))/(g) peaks at α = 45^°.
When linear aerodynamic drag (F_d = -kv) is introduced, horizontal velocity decays exponentially as vₓ(t) = v₀xe⁻(k/m)t, the descending branch becomes steeper than the ascending branch (Tartaglia’s ballistic curve), and the optimal launch angle for maximum range shifts below 45^°.
Everyday Analogy
When throwing a basketball or spraying water from a garden hose, 45^° gives the longest range in vacuum. In real air, aerodynamic drag slows the ball down and bends its path steeper at the end!
Interactive Question & Derivation
A projectile is launched at θ = 45^° to the horizontal in vacuum (c_d = 0). What is its kinetic energy Eₖ at the highest point (apex) of the trajectory compared to its initial kinetic energy Eₖ0 = (1)/(2)mv₀²?
- A: At the apex Eₖ = 0 because the body momentarily comes to a complete stop
- B: At the apex v_y = 0, but vₓ = v₀cos 45^°, so Eₖ = Eₖ0cos ² 45^° = (1)/(2)Eₖ0
- C: Kinetic energy at the apex equals the initial value Eₖ0
Study Hint: Compare the dashed vacuum parabola with the solid trajectory in air: what happens to the projectile’s mechanical energy and horizontal range when air drag acts against v?
Proof & Explanation: At the vertex of the parabola only the vertical velocity component vanishes (v_y = 0), while the horizontal component vₓ = v₀cos θ remains constant. Therefore, the minimum kinetic energy at the apex is Eₖ,min = (1)/(2)m(v₀cos θ)² = Eₖ0cos ²θ.
Key Equations & 3D Simulation Parameters
vₐpex = vₓ = v₀cos θ, Eₖ,min = Eₖ0cos ²θ
- Initial velocity v₀: 6 … 15 m/s (default: 11.5 m/s)
- Launch angle θ: 15 … 80 ° (default: 45 °)
- Initial height h₀: 0.2 … 4.5 m (default: 1.2 m)
- Air drag coeff. c_d: 0 … 0.15 kg/m (default: 0 kg/m)
- Projectile mass m: 0.5 … 5 kg (default: 1.5 kg)