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Projectile Launched at an Angle
A projectile launched at angle θ has vix = vi cos θ and viy = vi sin θ. Horizontal and vertical displacements and the equation of trajectory follow from independent component motions.
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3.2.2 Projectile Launched at an Angle
18. Resolution of Initial Velocity
Suppose a projectile is launched with initial velocity v_i at angle θ above the horizontal.
Horizontal Component
vix = vi cos θ
Vertical Component
viy = vi sin θ
19. Velocity Components at Time t
Horizontal Component
Since: ax = 0 Therefore: vx = vi cos θ
Vertical Component
Taking upward as positive: ay = −g Therefore: vy = viy − gt vy = vi sin θ − gt
20. Instantaneous Velocity and Direction
The velocity magnitude at time t is: v = √(vx² + vy²) v = √( (vi cos θ)² + (vi sin θ − gt)² ) Its direction is: tan φ = vyvx φ = tan⁻¹( vi sin θ − gtvi cos θ ) The sign of vy indicates whether the projectile is moving upward or downward.
21. Horizontal and Vertical Displacements
Horizontal Displacement
x = vix t x = vi cos θ · t
Vertical Displacement
y = viy t − ½gt² y = vi sin θ · t − ½gt²
22. Equation of Angled Trajectory
From horizontal motion: x = vi cos θ · t Therefore: t = xvi cos θ Substituting into the vertical equation: y = vi sin θ (xvi cos θ) − ½g (xvi cos θ)² y = x tan θ − gx² / (2 v_i² cos² θ) The presence of x² confirms that the ideal trajectory is a parabola.
29. Book-Based Solved Example: Angled Projection
Book-Based Solved Example: Angled Projection
A projectile is launched at 200 m s⁻¹ at an angle of 30°. Calculate its time of flight, maximum height and horizontal range. Take g = 9.8 m s⁻².
- vi = 200 m s⁻¹
- θ = 30°
- g = 9.8 m s⁻²
Time of Flight: T = 2 vi sin θg
T = 2(200) sin 30° / 9.8
T = 400(0.5) / 9.8
T = 20.41 s
Maximum Height: H = vi² sin² θ2g
H = (200)² (sin 30°)² / (2 × 9.8)
H = 40000(0.5)² / 19.6
H = 1000019.6
H = 510.2 m
Horizontal Range: R = vi² sin 2θ / g
R = (200)² sin 60° / 9.8
R = 40000(0.866) / 9.8
R = 3534.7 m
Final answer: T = 20.41 s; H = 510.2 m; R = 3534.7 m
32. Formula Summary — Angled Launch
Angled Launch
vix = vi cos θ viy = vi sin θ vx = vi cos θ vy = vi sin θ − gt H = vi² sin² θ2g T = 2 vi sin θg R = vi² sin 2θ / g Rmax = vfraci²g
Practice
Practice
Q1. Resolve an initial velocity into horizontal and vertical components.
Practice
Q2. Compare horizontal launch with angled launch.