Projectile Launched Horizontally

3.2.1 Projectile Launched Horizontally

8. Initial Conditions

Suppose a projectile is launched horizontally with initial speed vi. Its initial horizontal velocity is: vix = vi Its initial vertical velocity is: viy = 0 Horizontal acceleration is: ax = 0 Taking downward as positive, vertical acceleration is: ay = g

9. Horizontal Velocity

Since horizontal acceleration is zero: vx = vix + ax t vx = vi + (0)t vx = v_i The horizontal velocity remains constant when air resistance is ignored.

10. Vertical Velocity

The vertical motion begins from rest: viy = 0 Using the first equation of motion: vy = viy + gt vy = 0 + gt v_y = gt The downward vertical velocity increases with time.

11. Instantaneous Velocity

At any instant, the projectile has horizontal and vertical velocity components. Its velocity magnitude is: v = √(vx² + vy²) Substituting the components: v = √(vi² + g²t²) The direction below the horizontal is: tan φ = vyv_x tan φ = gt / vi φ = tan⁻¹(gt / vi)

12. Horizontal Displacement

Horizontal velocity remains constant. Therefore: x = vx t x = vi t

13. Vertical Displacement

Using: y = v_iy t + ½gt² Since: v_iy = 0 Therefore: y = (0)t + ½gt² y = ½gt²

14. Time of Fall From Height h

For a projectile falling through height h: h = ½gt² Multiplying by 2: 2h = gt² Dividing by g: t² = 2h / g Therefore: t = √(2h / g) The time of fall does not depend on the horizontal launch speed.

15. Horizontal Range From a Height

Horizontal distance is: x = vi t Substituting: t = √(2h / g) Therefore: x = vi √(2h / g)

16. Equation of Trajectory

From horizontal motion: x = vi t Therefore: t = xv_i For vertical motion: y = ½gt² Substituting t = xv_i: y = ½g (xv_i)² y = gx² / (2 vi²) Since y depends on x², the trajectory is parabolic.

17. Book-Based Solved Example: Horizontal Launch

32. Formula Summary — Horizontal Launch

Horizontal Launch

v_x = v_i v_y = gt x = v_i t y = ½gt² t = √(2h / g) x = v_i √(2h / g)