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Projectile Motion  IB DP Physics Study Notes

Projectile Motion  IB DP Physics Study Notes - 2025 Syllabus

Projectile Motion  IB DP Physics Study Notes

Projectile Motion  IB DP Physics Study Notes at  IITian Academy  focus on  specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand

  • Projectile Motion 
  • Effects of Fluid Resistance on Projectile Motion
  • Terminal Speed 

Standard level and higher level: 9 hours
Additional higher level: There is no additional higher level content

IB DP Physics 2025 -Study Notes -All Topics

IBDP Physics 2025 -Study Notes -All Topics

Projectile Motion 

Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration due to gravity.

Assumptions:

  • Air resistance is negligible.
  • Acceleration due to gravity \( g \) is constant (usually \( 9.8 \, \text{m/s}^2 \)).
  • Motion occurs in two perpendicular directions: horizontal and vertical.

Key Concepts:

1. Horizontal Motion

No acceleration in horizontal direction (assuming no air resistance), so horizontal velocity remains constant.
\( x = u_x t \), where \( u_x = u \cos \theta \)

2. Vertical Motion

Vertical motion is uniformly accelerated under gravity.
Equations of motion apply:

  • \( v_y = u_y – gt \)
  • \( y = u_y t – \frac{1}{2}gt^2 \)
  • \( v_y^2 = u_y^2 – 2g y \)

where \( u_y = u \sin \theta \)

3. Time of Flight (T)

Total time the projectile remains in the air.
For a projectile that lands at the same vertical level: \( T = \frac{2u \sin \theta}{g} \)

4. Maximum Height (H)

Maximum vertical displacement reached by the projectile:
\( H = \frac{u^2 \sin^2 \theta}{2g} \)

5. Horizontal Range (R)

Horizontal distance covered during entire flight:
\( R = \frac{u^2 \sin 2\theta}{g} \)

6. Path of Projectile

The trajectory is a parabola.
From horizontal and vertical equations, eliminate \( t \): \( y = x \tan \theta – \frac{g x^2}{2 u^2 \cos^2 \theta} \)

Important Notes:

  • Horizontal and vertical motions are independent.
  • Maximum range is achieved when \( \theta = 45^\circ \)
  • At maximum height, vertical velocity \( v_y = 0 \)

Example:

A ball is projected with an initial velocity of \( 20 \, \text{m/s} \) at an angle of \( 30^\circ \) above the horizontal.

  • (a) Calculate the time of flight.
  • (b) Calculate the maximum height reached.
  • (c) Calculate the horizontal range.
▶️ Answer/Explanation

Given:

\( u = 20 \, \text{m/s}, \theta = 30^\circ, g = 9.8 \, \text{m/s}^2 \)

(a) Time of flight

Time of flight: \( T = \frac{2u \sin \theta}{g} = \frac{2 \cdot 20 \cdot \sin 30^\circ}{9.8} \)
\( = \frac{40 \cdot 0.5}{9.8} = \frac{20}{9.8} \approx \boxed{2.04 \, \text{seconds}} \)

(b) Maximum height

Maximum height: \( H = \frac{u^2 \sin^2 \theta}{2g} = \frac{(20)^2 \cdot \sin^2 30^\circ}{2 \cdot 9.8} \)
\( = \frac{400 \cdot (0.5)^2}{19.6} = \frac{400 \cdot 0.25}{19.6} = \frac{100}{19.6} \approx \boxed{5.10 \, \text{m}} \)

(c) Horizontal Range

Range: \( R = \frac{u^2 \sin 2\theta}{g} = \frac{400 \cdot \sin 60^\circ}{9.8} \)
\( = \frac{400 \cdot 0.866}{9.8} = \frac{346.4}{9.8} \approx \boxed{35.35 \, \text{m}} \)

Example:

A ball is projected from the top of a \( 45 \, \text{m} \) high cliff with an initial speed of \( 25 \, \text{m/s} \) at an angle of \( 37^\circ \) above the horizontal.

Calculate:

  • (a) Time taken to hit the ground
  • (b) The horizontal distance from the base of the cliff where the ball lands
  • (c) The velocity of the ball just before it hits the ground
▶️ Answer/Explanation

Given:

Initial speed \( u = 25 \, \text{m/s} \)
Launch angle \( \theta = 37^\circ \)
Cliff height \( h = 45 \, \text{m} \)
\( u_x = u \cos \theta = 25 \cos 37^\circ = 25 \cdot 0.7986 \approx 19.97 \, \text{m/s} \)
\( u_y = u \sin \theta = 25 \cdot 0.6018 \approx 15.05 \, \text{m/s} \)

(a) Time of flight

Use vertical displacement equation: \( y = u_y t – \frac{1}{2}gt^2 \)
Set \( y = -45 \, \text{m} \) (falls below launch height), \( g = 9.8 \)
\( -45 = 15.05t – 4.9t^2 \)
Rewriting: \( 4.9t^2 – 15.05t – 45 = 0 \) Solve quadratic: \( t = \frac{15.05 \pm \sqrt{(-15.05)^2 + 4 \cdot 4.9 \cdot 45}}{2 \cdot 4.9} \)
\( t = \frac{15.05 \pm \sqrt{226.5 + 882}}{9.8} = \frac{15.05 \pm \sqrt{1108.5}}{9.8} \)
\( \sqrt{1108.5} \approx 33.29 \Rightarrow t = \frac{15.05 + 33.29}{9.8} \approx \frac{48.34}{9.8} \approx \boxed{4.93 \, \text{s}} \)

(b) Horizontal range

\( R = u_x \cdot t = 19.97 \cdot 4.93 \approx \boxed{98.5 \, \text{m}} \)

(c) Velocity before hitting ground

Final vertical velocity: \( v_y = u_y – g t = 15.05 – 9.8 \cdot 4.93 \approx 15.05 – 48.31 = -33.26 \, \text{m/s} \)
Horizontal velocity remains: \( v_x = 19.97 \, \text{m/s} \) Resultant speed: \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{(19.97)^2 + (-33.26)^2} \approx \sqrt{398.8 + 1106.2} = \sqrt{1505} \approx \boxed{38.79 \, \text{m/s}} \) Direction below horizontal: \( \tan \alpha = \frac{|v_y|}{v_x} = \frac{33.26}{19.97} \Rightarrow \alpha \approx \tan^{-1}(1.666) \approx \boxed{59.0^\circ \, \text{below horizontal}} \)

Qualitative Effects of Fluid Resistance on Projectile Motion

Fluid resistance (also called air resistance or drag) is a frictional force that opposes the motion of a body through a fluid (such as air). It depends on the object’s speed, shape, surface area, and the properties of the fluid.

General Impact:

In the presence of air resistance, the motion of the projectile deviates from ideal parabolic motion. The effects differ depending on whether the projectile is moving upward or downward.

Effect on Key Motion Quantities:

  • Time of Flight:

    Reduced — The upward motion is more affected due to opposing drag, causing the projectile to reach the peak sooner.

  • Trajectory:

    No longer symmetric or parabolic. The projectile falls more steeply than it rises. The path is skewed and shorter.

  • Velocity:

    Decreases faster during upward motion and increases slower during downward motion compared to ideal conditions.

  • Acceleration:

    The net acceleration is less than \( g \) when moving downward (because drag acts upward), and more than \( g \) when moving upward (because drag acts downward).

  • Range:

    Decreased — The horizontal velocity is continuously reduced due to drag, so the projectile travels a shorter horizontal distance.

Visual Summary:

  • Without air resistance: Symmetric parabolic path.
  • With air resistance: Asymmetric, steeper descent, shorter range, curved trajectory.

Important Notes:

  • Air resistance has a larger effect on lighter and larger-surface-area objects.
  • At high speeds, drag force becomes proportional to \( v^2 \).

Example:

A student throws a ball with a speed of \( 20 \, \text{m/s} \) at an angle of \( 45^\circ \) above the horizontal. Describe and compare the motion of the ball:

  • (a) Ignoring air resistance
  • (b) Considering air resistance qualitatively
▶️ Answer/Explanation

(a) Ignoring air resistance

 The path is a symmetric parabola.
 Time of flight: \( T = \frac{2u \sin \theta}{g} = \frac{2 \cdot 20 \cdot \sin 45^\circ}{9.8} \approx 2.89 \, \text{s} \)
 Maximum height: \( H = \frac{u^2 \sin^2 \theta}{2g} = \frac{(20^2)(0.5)}{2 \cdot 9.8} \approx 10.2 \, \text{m} \)
 Horizontal range: \( R = \frac{u^2 \sin 2\theta}{g} = \frac{400 \cdot 1}{9.8} \approx 40.8 \, \text{m} \)
 Acceleration is \( 9.8 \, \text{m/s}^2 \) downward at all times.

(b) With air resistance (qualitative)

 The path is asymmetric and not a perfect parabola.
 Time of flight is shorter than 2.89 s because upward velocity is reduced more quickly.
 Maximum height is less than 10.2 m because air drag reduces vertical motion.
 Range is shorter than 40.8 m because horizontal speed is reduced by drag.
 Acceleration: Greater than \( g \) during upward motion (drag + gravity), less than \( g \) on way down (drag opposes gravity).
 Final speed before landing is less than the initial speed due to energy lost to air drag.

Conclusion:

Air resistance causes the projectile to:

  • Spend less time in the air
  • Reach a lower maximum height
  • Land at a shorter horizontal distance
  • Have a non-symmetric trajectory

Terminal Speed 

Terminal speed (or terminal velocity) is the maximum constant speed a falling object reaches when the drag force equals the weight of the object, resulting in zero net acceleration.

Forces Acting:

  • Weight (mg): Downward force due to gravity
  • Drag force (FD): Upward resistive force from the fluid (e.g. air)

Condition for Terminal Speed:

When falling:

  • Initially: \( F_{\text{net}} = mg – F_D > 0 \) → acceleration downward
  • As speed increases, \( F_D \) increases
  • Eventually: \( F_D = mg \) → \( F_{\text{net}} = 0 \), acceleration = 0
  • Object continues at constant terminal speed

Equation for Drag Force:

At high speeds in fluids, drag force is proportional to the square of velocity:
\( F_D = \frac{1}{2} C \rho A v^2 \), where:

  • \( C \): Drag coefficient (depends on shape)
  • \( \rho \): Fluid density
  • \( A \): Cross-sectional area
  • \( v \): Speed of object

Expression for Terminal Speed:

Equating drag force to weight:
\( mg = \frac{1}{2} C \rho A v_T^2 \)
Solving for \( v_T \): \( v_T = \sqrt{ \frac{2mg}{C \rho A} } \)

Factors Affecting Terminal Speed:

  • ↑ Mass → ↑ Terminal speed
  • ↑ Area → ↓ Terminal speed
  • ↑ Drag coefficient (less aerodynamic shape) → ↓ Terminal speed
  • ↑ Fluid density → ↓ Terminal speed

Applications:

  • Skydiving: Terminal speed is reached during free-fall
  • Parachutes: Increase drag, reducing terminal speed
  • Raindrops: Reach terminal speed before hitting the ground

Graphical Understanding:

A velocity-time graph of a falling object with air resistance shows:

  • Increasing velocity with decreasing acceleration
  • Horizontal asymptote at \( v_T \)

Example:

A small spherical object of mass \( 0.15 \, \text{kg} \) and cross-sectional area \( 0.003 \, \text{m}^2 \) is dropped from rest in air. The drag coefficient is \( 0.47 \), and the density of air is \( 1.2 \, \text{kg/m}^3 \).

Calculate the terminal speed of the object.

▶️ Answer/Explanation

Given:

\( m = 0.15 \, \text{kg} \)
\( A = 0.003 \, \text{m}^2 \)
\( C = 0.47 \)
\( \rho = 1.2 \, \text{kg/m}^3 \)
\( g = 9.8 \, \text{m/s}^2 \)

Formula for terminal speed:

\( v_T = \sqrt{ \frac{2mg}{C \rho A} } \)

Substitute values:

\( v_T = \sqrt{ \frac{2 \cdot 0.15 \cdot 9.8}{0.47 \cdot 1.2 \cdot 0.003} } \)
\( v_T = \sqrt{ \frac{2.94}{0.001692} } \)
\( v_T = \sqrt{1738.68} \approx \boxed{41.7 \, \text{m/s}} \)

Conclusion:

The object will accelerate until it reaches a terminal speed of approximately \( 41.7 \, \text{m/s} \), after which it continues falling at constant speed.

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