The path of an object projected horizontally or thrown upward at an angle is a parabola. The key is the independence of horizontal and vertical motion.
Basic Knowledge
Projectile motion is the combination of horizontal motion (uniform linear motion) and vertical motion (free fall and uniformly accelerated motion).
Horizontal projection (initial speed v₀ given horizontally):
・Horizontal: x = v₀t (acceleration = 0)
・Vertical: y = ½gt² (initial velocity = 0, downward positive)
・Velocity: vₓ = v₀、vᵧ = gt
Oblique projection (initial speed v₀ thrown upward at angle θ):
・Horizontal: x = v₀cosθ · t
・Vertical: y = v₀sinθ · t − ½gt²
・Horizontal range (returning to the horizontal plane): R = v₀²sin2θ / g
・Velocity at the highest point: only the horizontal component v₀cosθ (vertical component = 0)
Horizontal projection: x = v₀t、y = ½gt²
Range of oblique projection: R = v₀²sin2θ / g
Angle for maximum range: θ = 45° (maximum when sin90° = 1)
Velocity at the highest point: vᵧ = 0 → velocity is v₀cosθ (horizontal only)
Time of flight: t = 2v₀sinθ / g (from the condition y = 0)
Further Study (Background and Significance)
Because horizontal and vertical motion are completely independent in projectile motion, the proper method is to set up separate equations of motion for each coordinate axis. This is called the independence of motion, an important principle confirmed experimentally by Galileo.
From R = v₀²sin2θ / g, the range is greatest at θ = 45°, and the ranges at θ and 90° − θ (complementary angles) are equal (for example, the ranges at 30° and 60° are the same). This property is frequently tested.
At the highest point, vertical velocity becomes vᵧ = 0, but horizontal velocity remains v₀cosθ. At this instant, acceleration is still g (vertically downward) (zero velocity does not mean zero acceleration).
For actual projectiles and athletic throws, air resistance makes the optimum angle smaller than 45°, but in a vacuum it is exactly 45°.
- Horizontal: uniform linear motion (zero acceleration)
- Vertical: uniformly accelerated motion (acceleration g downward)
- Solve the two directions independently
- At the highest point, vᵧ = 0 (horizontal velocity remains)
- Maximum range at θ = 45°
- The ranges at θ and (90°−θ) are equal
- Set up simultaneous equations for x and y using time t
Points to Note (Common Confusions)
① Do not mistake the velocity for zero at the highest point. Only the vertical velocity is zero; horizontal velocity remains. ② Acceleration is still g at the highest point (gravity always acts downward). ③ “Horizontal projection” is the special case θ = 0. Understand the unified formulas. ④ Decide at the beginning whether upward or downward is positive, and use the sign convention consistently.
Practice
- An object projected horizontally at an initial speed of 20 m/s: find its horizontal and vertical positions after 2.0 s (g = 10 m/s²).
- Find the time of flight when an object is projected obliquely with initial speed v₀ at angle θ.
- What projection angle gives the maximum range, and what other angle gives the same range (when θ = 30°)?