A1-4. Projectile Motion

Acceleration Due To Gravity

An object that falls freely will accelerate towards the Earth because of the force of gravity acting on it. The size of this acceleration does not depend on mass—meaning a feather and a bowling ball accelerate at the exact same rate in a vacuum.

Gravitational Field Strength (g): Near the surface of the Earth, the gravitational field strength is 9.81 N/kg. This is also the acceleration a free-falling object experiences. In the equations of motion, a = g = 9.81 m/s².

Mass vs. Weight:

  • Mass (m): A property that tells us how much matter an object is made of. Measured in kilograms (kg).
  • Weight (W): A force caused by gravity acting on a mass. Measured in Newtons (N).
$$W = mg$$

Example 1: Free Fall

Problem: Bob falls freely from rest from a high platform. He reaches the water surface after 2.0 s. Assuming $g = 10 \text{ m/s}^2$, determine:

  1. His velocity upon hitting the water
  2. The height of the platform from the water

Solution:

  • a) Velocity upon impact ($v$):
    Taking downwards as positive: $u = 0$, $a = g = 10 \text{ m/s}^2$, $t = 2.0 \text{ s}$
    $$v = gt \implies v = 10 \times 2.0 = \mathbf{20 \text{ m/s}}$$ (downwards)
  • b) Height of platform ($H$):
    $$H = \dfrac{1}{2}gt^2 \implies H = \dfrac{1}{2}(10)(2.0)^2 = 5 \times 4 = \mathbf{20 \text{ m}}$$

Terminal Velocity

If an object is dropped from a plane, it initially accelerates due to its weight. As its velocity increases, the air resistance (drag forces) acting on the object also increases.

Eventually, the upward air resistance will perfectly balance the downward weight of the object. This means there is no overall net force, and therefore no acceleration (a = 0). The object stops accelerating and has reached its maximum, constant speed called Terminal Velocity.

Independence of Motion

⚠ The Golden Rule of Projectiles

The most important concept in projectiles is that horizontal and vertical motions are completely independent of each other.

If you drop a ball vertically from rest, and simultaneously fire another ball horizontally from the exact same height, they will hit the ground at the exact same time. The horizontal velocity has absolutely no effect on the vertical rate of falling.

Dropped Fired Horizontally Hit ground simultaneously

Horizontal and Vertical Breakdown

Component Horizontal Motion (x-axis) Vertical Motion (y-axis)
Acceleration ($a$) 0 m/s² (No horizontal forces) -9.81 m/s² (Gravity pulling down)
Velocity ($v$) Velocity is constant. It covers the same horizontal distance every second. Velocity changes constantly. It accelerates on the way down, or decelerates on the way up.
Equations to Use $v_x = \dfrac{x}{t}$ Use the 4 SUVAT equations.

Horizontal Projectile Motion

When an object is launched perfectly horizontally with an initial velocity $v_{0x}$, it has no initial vertical velocity ($v_{0y} = 0$). Taking the downward direction as positive:

  • Horizontal Position: $x = v_{0x}t$
  • Vertical Position: $y = \dfrac{1}{2}gt^2$
  • Trajectory Equation (Parabola): By substituting $t = \dfrac{x}{v_{0x}}$ into the vertical equation, the path follows:
    $$y = \dfrac{g}{2{v_{0x}}^2}x^2$$
  • Velocity at any time $t$: $v = \sqrt{{v_x}^2 + {v_y}^2}$

Example 2: Horizontal Projectile

Problem: Kevin is hiking and encounters two rocks with a horizontal distance of 4.0 m between them. The rock he is standing on is 2.5 m higher than the target rock. If $g = 9.8 \text{ m/s}^2$ and he jumps perfectly horizontally, determine:

  1. The time it takes to land on the lower rock
  2. The minimum initial velocity required to safely reach the rock

Solution:

  • a) Time to land ($t$):
    Looking only at the vertical motion: $y = 2.5 \text{ m}$, $v_{0y} = 0$, $a_y = 9.8 \text{ m/s}^2$
    $$y = \dfrac{1}{2}gt^2 \implies 2.5 = \dfrac{1}{2}(9.8)t^2 \implies t^2 = \dfrac{2.5}{4.9} \approx 0.51 \implies t = \mathbf{0.71 \text{ s}}$$
  • b) Minimum initial velocity ($v_{0x}$):
    Looking at horizontal motion: $x = 4.0 \text{ m}$, $t = 0.71 \text{ s}$
    $$x = v_{0x}t \implies 4.0 = v_{0x}(0.71) \implies v_{0x} = \dfrac{4.0}{0.71} = \mathbf{5.6 \text{ m/s}}$$

Oblique Projectile Motion

An object kicked or thrown into the air at an angle will follow a full parabolic path. If launched with an initial velocity $v_0$ at an angle $\theta$ above the horizontal, we must resolve this vector into two components. As the object travels, its horizontal velocity ($v_x$) remains constant, while its vertical velocity ($v_y$) changes due to gravity.

Velocity Decomposition Along the Trajectory

The diagram below illustrates how the velocity vector ($\vec{v}$) is always tangent to the parabolic path. It can be broken down into $v_x$ (red, constant) and $v_y$ (green, changing).

x (m) y (m) v₀ v₀ₓ v₀y θ v₁ v₁ₓ v₁y v₂ = v₂ₓ (v₂y = 0) v₃ v₃ₓ v₃y v₄ v₄ₓ v₄y g x y v₀ v₀cosθ v₀sinθ θ H R

Kinematic Equations for Angled Projectiles

Taking upward as the positive direction ($a_y = -g$):

  • Initial Components: $v_{0x} = v_0 \cos\theta$ and $v_{0y} = v_0 \sin\theta$
  • Horizontal Position: $x = (v_0 \cos\theta)t$
  • Vertical Position: $y = (v_0 \sin\theta)t - \dfrac{1}{2}gt^2$
  • Trajectory Equation: $y = x \tan\theta - \dfrac{g}{2(v_0 \cos\theta)^2}x^2$

Example 3: Oblique Projectile

Problem: Football player Chris kicks a ball from the field with an initial velocity of $20 \text{ m/s}$ at an angle of $30^\circ$ above the horizontal. Assuming $g = 10 \text{ m/s}^2$ and ignoring air resistance, determine:

  1. The maximum height reached by the ball
  2. The total time the ball is in the air (Time of Flight)
  3. The horizontal distance covered (Range)

Solution:

Taking upward and rightward as positive ($a_y = -10 \text{ m/s}^2$, $a_x = 0 \text{ m/s}^2$):

Step 1: Resolve the initial velocity

  • Vertical initial velocity: $v_{0y} = v_0 \sin(30^\circ) = 20 \times 0.5 = 10 \text{ m/s}$
  • Horizontal initial velocity: $v_{0x} = v_0 \cos(30^\circ) = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3} \text{ m/s}$
  • a) Maximum Height ($H$):
    Concept: At the very top of the parabolic trajectory, the ball stops moving upwards, meaning its instantaneous vertical velocity is zero ($v_y = 0$).
    Using the equation $v_y^2 = {v_{0y}}^2 + 2a_y y$ (where $y = H$): $$0^2 = 10^2 + 2(-10)H = 100 - 20H \implies 20H = 100 \implies H = \mathbf{5 \text{ m}}$$
  • b) Time of Flight ($t$):
    Concept: The ball lands back on the ground when its net vertical displacement is zero ($y = 0$).
    Using the equation $y = v_{0y}t + \dfrac{1}{2}a_y t^2$: $$0 = 10t + \dfrac{1}{2}(-10)t^2 = 10t - 5t^2 = 5t(2 - t)$$ Since $t = 0$ represents the launch moment, the landing time is $t = \mathbf{2 \text{ s}}$.
  • c) Horizontal Range ($R$):
    Concept: The ball travels horizontally at a constant speed for its entire time of flight.
    Using the equation $x = v_{0x}t$ (where $x = R$ and $t = 2 \text{ s}$): $$R = 10\sqrt{3} \times 2 = 20\sqrt{3} \approx \mathbf{34.6 \text{ m}}$$