Grade 12 Physics – Two-Dimensional Motion (Introduction)

Projectile Motion – Grade 12 Physics

2.1 Projectile Motion

Projectile: A thrown, fired, or released object that moves only under the influence of gravitational force. The projectile acceleration is g = 9.8 m/s² downward.

Anyone who has observed the motion of a ball kicked by a football player (Figure 2.1b) has observed projectile motion. The ball moves in a curved path and returns to the ground. Other examples include a cannonball fired from a cannon, a bullet fired from a gun, the flight of a golf ball, and a jet of water escaping a hose.

Projectile Motion Examples
Figure 2.1: (a) A ball thrown horizontally (b) A football kicked in a game

Assumptions in Projectile Motion Analysis

Projectile motion is simple to analyze if we make three key assumptions:

  1. The free-fall acceleration is constant over the range of motion and always directed downward (g = 9.8 m/s²).
  2. The effect of air resistance is negligible.
  3. The horizontal velocity is constant because the acceleration has no horizontal component.

With these assumptions, the path of a projectile, called its trajectory, is a parabola.

Independence of Horizontal and Vertical Motions

The horizontal and vertical components of a projectile’s motion are completely independent of each other and can be handled separately, with time t as the common variable for both components.

Key Insight: This independence means we can analyze the horizontal motion using constant velocity equations and the vertical motion using constant acceleration equations (free fall), then combine the results using time as the connecting variable.

Horizontal Projection

In horizontal projection, the projectile is launched horizontally from a certain height. Its initial vertical velocity is zero, and it possesses only horizontal velocity at the beginning. As time progresses, gravity gives it a vertical velocity component.

Horizontal Projectile Motion
Figure 2.2: The motion of a ball projected horizontally

Equations for Horizontal Motion

Since there’s no horizontal acceleration, the horizontal velocity remains constant:

\( v_x = v_{0x} \) (constant)

The horizontal distance traveled at time t is:

\( \Delta x = v_{0x} t \)      (Equation 2.1)

Assume that an airplane flying horizontally drops a package to a remote village. What kind of motion is performed by the package? Draw the trajectory of the package. As the package hits the ground at the village, where is the aircraft?

Equations for Vertical Motion

The vertical motion is constant accelerated motion (free fall). The final vertical velocity after time t is:

\( v_y = v_{0y} + gt \)      (Equation 2.2)

Since the initial vertical velocity is zero (\(v_{0y} = 0\)):

\( v_y = gt \)

The vertical displacement is given by:

\( \Delta y = v_{0y}t + \frac{1}{2}gt^2 \)      (Equation 2.3)

With \(v_{0y} = 0\):

\( \Delta y = \frac{1}{2}gt^2 \)

Sign Convention: In vertical motion equations, upward is positive (+) and downward is negative (-). Since gravity acts downward, g is negative (-9.8 m/s²) in these equations.

Time of Flight

The time of flight is the time taken for the projectile to hit the ground:

\( \Delta y = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2\Delta y}{|g|}} \)

Range

The range is the maximum horizontal distance traveled:

\( R = v_{0x} \sqrt{\frac{2\Delta y}{|g|}} \)

Activity 2.1: Place two tennis balls at the edge of a tabletop. Sharply snap one ball horizontally off the table with one hand while gently tapping the second ball off with your other hand. Measure the height (y) of the table and the horizontal distance (R) from the table’s edge to the landing location.

Determine: (a) The time of flight of both tennis balls. Explain your result. (b) The initial horizontal velocity when they leave the table.

Example 2.1: A rifle is aimed horizontally at a target 30m away. The bullet hits 2 cm below the aiming point. (a) What is the bullet’s time of flight? (b) What is the initial velocity? (Use g = 10 m/s²)

Solution:
Given: \(\Delta x = 30 \text{ m}\), \(\Delta y = -0.02 \text{ m}\) (downward), \(g = -10 \text{ m/s}^2\)

(a) Using vertical motion equation:
\(\Delta y = \frac{1}{2}gt^2\)
\(-0.02 = \frac{1}{2}(-10)t^2\)
\(0.02 = 5t^2\)
\(t^2 = 0.004\)
\(t = \sqrt{0.004} = 0.063 \text{ s}\)

(b) Using horizontal motion equation:
\(\Delta x = v_{0x}t\)
\(30 = v_{0x} \times 0.063\)
\(v_{0x} = \frac{30}{0.063} \approx 476 \text{ m/s}\)

Inclined Projectile Motion

This occurs when an object is projected with initial velocity \(v_0\) at an angle \(\theta\) to the horizontal (Figure 2.4). The initial velocity can be resolved into horizontal and vertical components.

Inclined Projectile Motion
Figure 2.4: Inclined projectile motion

Velocity Components

Horizontal: \( v_{0x} = v_0 \cos \theta \)
Vertical: \( v_{0y} = v_0 \sin \theta \)

The horizontal velocity remains constant throughout the flight:

\( v_x = v_0 \cos \theta \)

The vertical velocity changes with time due to gravity:

\( v_y = v_0 \sin \theta + gt \)      (Equation 2.4)

Displacement Equations

Horizontal displacement at time t:

\( \Delta x = v_0 \cos \theta \cdot t \)      (Equation 2.5)

Vertical displacement at time t:

\( \Delta y = v_0 \sin \theta \cdot t + \frac{1}{2}gt^2 \)      (Equation 2.6)

Time to Reach Maximum Height

At maximum height, vertical velocity \(v_y = 0\):

\( 0 = v_0 \sin \theta + gt \implies t = -\frac{v_0 \sin \theta}{g} \)

Since g is negative, this gives a positive time:

\( t_{\text{max height}} = \frac{v_0 \sin \theta}{|g|} \)

Time of Flight

For projectiles that land at the same elevation as launch (\(\Delta y = 0\)):

\( 0 = v_0 \sin \theta \cdot t + \frac{1}{2}gt^2 \)
\( t_{\text{total}} = \frac{2v_0 \sin \theta}{|g|} \)

Note: This equation only applies when launch and landing elevations are equal. For different elevations, solve the quadratic equation for \(\Delta y = v_0 \sin \theta \cdot t + \frac{1}{2}gt^2\).

Horizontal Range and Maximum Height

Range (R): The maximum horizontal distance when \(\Delta y = 0\):

\( R = v_0 \cos \theta \cdot t_{\text{total}} = v_0 \cos \theta \cdot \frac{2v_0 \sin \theta}{|g|} = \frac{v_0^2 \sin 2\theta}{|g|} \)

Maximum Height (H): When \(v_y = 0\):

\( H = \frac{v_0^2 \sin^2 \theta}{2|g|} \)

Key Observations:

  • Range is maximum at \(\theta = 45^\circ\) (since \(\sin 90^\circ = 1\))
  • Projectiles launched at complementary angles (\(\theta\) and \(90^\circ – \theta\)) have the same range
  • The trajectory is symmetric about the maximum height point when launch and landing elevations are equal
Projectile Trajectories
Figure 2.5: Trajectories of projectiles (a) Effect of initial velocity (b) Effect of launch angle

Discussion Question 2.4:

  1. A projectile’s horizontal range equals three times its maximum height. What is the launch angle?
  2. At maximum height, which statement is true about velocity and acceleration?
  3. One ball is thrown horizontally, another is dropped from the same height. Which hits first? Explain.

Relation Between Range and Maximum Height

From the equations for R and H:

\( \frac{H}{R} = \frac{v_0^2 \sin^2 \theta / 2|g|}{v_0^2 \sin 2\theta / |g|} = \frac{\sin^2 \theta}{2 \cdot 2 \sin \theta \cos \theta} = \frac{\sin \theta}{4 \cos \theta} = \frac{\tan \theta}{4} \)
\( H = \frac{R \tan \theta}{4} \)

Example 2.3: A football player kicks a ball at 37° with horizontal. Initial velocity is 40 m/s. (a) Find maximum height. (b) Find horizontal range. (Use g = 10 m/s²)

Solution:
Given: \(v_0 = 40 \text{ m/s}\), \(\theta = 37^\circ\), \(g = 10 \text{ m/s}^2\)
\(\sin 37^\circ \approx 0.6\), \(\cos 37^\circ \approx 0.8\)

(a) Maximum height:
\(H = \frac{v_0^2 \sin^2 \theta}{2g} = \frac{40^2 \times (0.6)^2}{2 \times 10} = \frac{1600 \times 0.36}{20} = \frac{576}{20} = 28.8 \text{ m}\)

(b) Horizontal range:
\(R = \frac{v_0^2 \sin 2\theta}{g} = \frac{40^2 \times \sin 74^\circ}{10} = \frac{1600 \times 0.961}{10} \approx 153.8 \text{ m}\)

Special Cases and Problem Solving

Projectile motion problems often involve:

  • Projectiles landing at different elevations
  • Projectiles hitting targets at specific points
  • Projectiles launched from moving platforms

Example 2.4: A ball is kicked toward a wall 24m away with initial speed 25 m/s at 53° above horizontal. (a) Time to reach wall? (b) Height when hitting wall? (c) Velocity components at impact? (d) Resultant velocity?

Solution:
Given: \(\Delta x = 24 \text{ m}\), \(\theta = 53^\circ\), \(v_0 = 25 \text{ m/s}\)
\(\sin 53^\circ \approx 0.8\), \(\cos 53^\circ \approx 0.6\)

(a) Time to reach wall:
\(\Delta x = v_0 \cos \theta \cdot t\)
\(24 = 25 \times 0.6 \times t\)
\(24 = 15t\)
\(t = 1.6 \text{ s}\)

(b) Height at impact:
\(\Delta y = v_0 \sin \theta \cdot t + \frac{1}{2}gt^2\)
\(\Delta y = 25 \times 0.8 \times 1.6 + \frac{1}{2}(-10)(1.6)^2\)
\(\Delta y = 32 – 12.8 = 19.2 \text{ m}\)

(c) Velocity components:
\(v_x = v_0 \cos \theta = 25 \times 0.6 = 15 \text{ m/s}\)
\(v_y = v_0 \sin \theta + gt = 25 \times 0.8 + (-10)(1.6) = 20 – 16 = 4 \text{ m/s}\)

(d) Resultant velocity:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 4^2} = \sqrt{225 + 16} = \sqrt{241} \approx 15.5 \text{ m/s}\)

Activity 2.4: Use conservation of mechanical energy to investigate inclined projection. Calculate the speed of a ball leaving a track, then predict where it will land using projectile motion equations.

Discussion Question 2.5:

  1. Is there any point in the trajectory where velocity and acceleration are (a) perpendicular? (b) parallel?
  2. Which statements about projectile motion are true?
  3. Compare projectile motion on Earth vs. Moon (different gravity).

Key Concepts Summary

  • Projectile motion can be analyzed by separating horizontal (constant velocity) and vertical (constant acceleration) components
  • Time is the common variable connecting both motions
  • Horizontal velocity remains constant throughout the motion
  • Vertical motion follows free-fall equations with constant acceleration g
  • Maximum range occurs at 45° launch angle (for same elevation)
  • Complementary launch angles (θ and 90°-θ) yield the same range
  • Trajectory is parabolic under constant gravity with no air resistance

Content adapted from: Grade 12 Physics New Curriculum (2023), Federal Democratic Republic of Ethiopia, Ministry of Education

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