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Projectile Launched Horizontally
A projectile launched horizontally has v_iy = 0. Horizontal velocity remains constant. Vertical displacement is y = ½gt². Time of fall from height h is t = √(2h/g).
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Recommended for Class 11
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 = vyvx 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 = viy t + ½gt² Since: viy = 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 = xvi For vertical motion: y = ½gt² Substituting t = xvi: y = ½g (xvi)² y = gx² / (2 vi²) Since y depends on x², the trajectory is parabolic.
17. Book-Based Solved Example: Horizontal Launch
Book-Based Solved Example: Horizontal Launch
A projectile is launched horizontally at 12.0 m s⁻¹ from the top of a 24.6 m building. Determine its horizontal displacement. Take g = 9.8 m s⁻².
- vi = 12.0 m s⁻¹
- h = 24.6 m
- g = 9.8 m s⁻²
Finding Time: h = ½gt²
24.6 = ½(9.8)t²
24.6 = 4.9t²
t² = 24.64.9
t² = 5.02
t = 2.24 s
Finding Horizontal Displacement: x = v_i t
x = (12.0)(2.24)
x = 26.88 m
Final answer: x ≈ 26.9 m
32. Formula Summary — Horizontal Launch
Horizontal Launch
vx = vi vy = gt x = vi t y = ½gt² t = √(2h / g) x = vi √(2h / g)
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Practice
Q1. A ball is launched horizontally at 15 m s⁻¹ from a height of 20 m. Find its time of fall and horizontal distance using g = 10 m s⁻².