IB — physicsMedium3 min read

Projectile Motion

Independent horizontal and vertical components; range; trajectory; air resistance effects

Tutor AI Team
Tutor AI Team

# Projectile Motion — IB Physics

Projectile motion combines constant horizontal velocity with uniformly accelerated vertical motion. The key insight: horizontal and vertical components are independent.

1. Resolving the Initial Velocity

For launch angle θ\theta and initial speed uu:

  • Horizontal: ux=ucos⁡θu_x = u\cos\theta
  • Vertical: uy=usin⁡θu_y = u\sin\theta

2. Equations of Motion

Horizontal (no acceleration)

x=uxt=ucos⁡θ⋅tx = u_x t = u\cos\theta \cdot t

Vertical (a=−ga = -g, taking up as positive)

vy=uy−gt=usin⁡θ−gtv_y = u_y - gt = u\sin\theta - gt y=uyt−12gt2y = u_y t - \frac{1}{2}gt^2

3. Key Results (Launching from Ground Level)

Time of flight: T=2usin⁡θgT = \frac{2u\sin\theta}{g}

Maximum height: H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g}

Range: R=u2sin⁡2θgR = \frac{u^2\sin 2\theta}{g}

Maximum range occurs at θ=45°\theta = 45°.

4. Launching Horizontally

For an object launched horizontally from height hh with speed uu:

  • ux=uu_x = u, uy=0u_y = 0
  • Time to fall: h=12gt2h = \frac{1}{2}gt^2 → t=2h/gt = \sqrt{2h/g}
  • Horizontal distance: x=utx = ut

5. Effect of Air Resistance

  • Reduces range and maximum height
  • Path is no longer symmetric
  • Object reaches a lower maximum height
  • Falls more steeply

Worked Example: Horizontal Launch

Problem

A ball rolls off a 1.25 m high table at 3 m/s. Find where it lands.

t=2h/g=2(1.25)/9.81=0.505t = \sqrt{2h/g} = \sqrt{2(1.25)/9.81} = 0.505 s x=ut=3(0.505)=1.52x = ut = 3(0.505) = 1.52 m from the base of the table.

Solution

Worked Example: Angled Launch

Problem

A ball is launched at 30 m/s at 60° from ground level.

ux=30cos⁡60°=15u_x = 30\cos60° = 15 m/s; uy=30sin⁡60°=26.0u_y = 30\sin60° = 26.0 m/s T=2(26.0)/9.81=5.30T = 2(26.0)/9.81 = 5.30 s H=(30)2sin⁡260°/(2×9.81)=900(0.75)/19.62=34.4H = (30)^2\sin^2 60°/(2 \times 9.81) = 900(0.75)/19.62 = 34.4 m R=(30)2sin⁡120°/9.81=900(0.866)/9.81=79.4R = (30)^2\sin120°/9.81 = 900(0.866)/9.81 = 79.4 m

Solution

Worked Example: Velocity at Impact

Problem

For Example 2, find the speed when the ball returns to ground.

By symmetry: vx=15v_x = 15 m/s, vy=−26.0v_y = -26.0 m/s v=152+26.02=225+676=30v = \sqrt{15^2 + 26.0^2} = \sqrt{225 + 676} = 30 m/s (same as launch speed!)

Solution

7. Practice Questions

    1. A stone is thrown horizontally at 8 m/s from a cliff 80 m high. Find the horizontal distance and speed on impact. (5 marks)
    1. A ball is kicked at 20 m/s at 45°. Calculate the range and maximum height. (4 marks)
    1. Explain why 30° and 60° give the same range for the same launch speed. (2 marks)

    Answers

    1. t=2(80)/9.81=4.04t = \sqrt{2(80)/9.81} = 4.04 s. x=8(4.04)=32.3x = 8(4.04) = 32.3 m. vy=gt=39.6v_y = gt = 39.6 m/s. v=64+1568=40.4v = \sqrt{64 + 1568} = 40.4 m/s.
    1. R=202sin⁡90°/9.81=400/9.81=40.8R = 20^2\sin90°/9.81 = 400/9.81 = 40.8 m. H=202sin⁡245°/(2×9.81)=400(0.5)/19.62=10.2H = 20^2\sin^2 45°/(2 \times 9.81) = 400(0.5)/19.62 = 10.2 m.

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Summary

  • Resolve velocity into horizontal and vertical components
  • Horizontal: constant velocity (ax=0a_x = 0)
  • Vertical: uniformly accelerated (ay=−ga_y = -g)
  • T=2usin⁡θ/gT = 2u\sin\theta/g; H=u2sin⁡2θ/(2g)H = u^2\sin^2\theta/(2g); R=u2sin⁡2θ/gR = u^2\sin 2\theta/g
  • Max range at 45°; complementary angles give same range
Tags:
#ib#physics#mechanics#projectile-motion#trajectory#kinematics

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