Projectile Launched at an Angle

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

v_ix = v_i cos θ

Vertical Component

v_iy = v_i sin θ

19. Velocity Components at Time t

Horizontal Component

Since: a_x = 0 Therefore: v_x = v_i cos θ

Vertical Component

Taking upward as positive: ay = −g Therefore: vy = viy − gt vy = v_i 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 = vyv_x φ = tan⁻¹( v_i sin θ − gtv_i cos θ ) The sign of vy indicates whether the projectile is moving upward or downward.

21. Horizontal and Vertical Displacements

Horizontal Displacement

x = v_ix t x = v_i cos θ · t

Vertical Displacement

y = v_iy t − ½gt² y = v_i sin θ · t − ½gt²

22. Equation of Angled Trajectory

From horizontal motion: x = vi cos θ · t Therefore: t = xv_i cos θ Substituting into the vertical equation: y = vi sin θ (xv_i cos θ) − ½g (xv_i 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

32. Formula Summary — Angled Launch

Angled Launch

v_ix = v_i cos θ v_iy = v_i sin θ v_x = v_i cos θ v_y = v_i sin θ − gt H = v_i² sin² θ2g T = 2 v_i sin θg R = v_i² sin 2θ / g R_max = v_fraci²g