3.11 Ideal Projectile Motion | Translatory Motion | Physics 11

For the purpose of basic analysis, we study an ideal projectile. This model makes several simplifying assumptions:

Air resistance is negligible.
The acceleration due to gravity, g, is constant (approximately $9.8 ms^{- 2}$ ) and directed downwards.
The rotation of the Earth does not affect the motion.

Under these conditions, the trajectory of a projectile is a parabola.

Figure 1: Trajectory of a projectile showing parabolic path

Analysis of Motion: Separating the Components

The key to solving projectile motion problems is to analyze the horizontal and vertical components of motion separately. This is a practical application of Rectangular Components Of A Vector→.

Horizontal Motion

Acceleration ( $a_{x}$ ): Since we ignore air resistance, there are no horizontal forces acting on the projectile. Therefore, the horizontal acceleration is zero. $a_{x} = 0$
Velocity ( $v_{x}$ ): Because the acceleration is zero, the horizontal component of the velocity is constant throughout the flight. $v_{f x} = v_{i x} = constant$
Displacement ( $x$ ): The horizontal displacement (range) is simply the constant horizontal velocity multiplied by time. $x = v_{i x} t$

Vertical Motion

Acceleration ( $a_{y}$ ): The only force acting vertically is gravity. Therefore, the vertical acceleration is constant and directed downwards. $a_{y} = - g$
Velocity ( $v_{y}$ ): The vertical velocity changes continuously due to gravity. It can be found using the first equation of motion. $v_{f y} = v_{i y} - g t$
Displacement ( $y$ ): The vertical displacement (height) changes over time and can be found using the second equation of motion. $y = v_{i y} t - \frac{1}{2} g t^{2}$

Initial Velocity Components

If a projectile is launched with an initial velocity $v_{i}$ at an angle $θ$ above the horizontal, we must first resolve this velocity into its x and y components.

Horizontal Initial Velocity ( $v_{i x}$ ): $v_{i x} = v_{i} cos θ$
Vertical Initial Velocity ( $v_{i y}$ ): $v_{i y} = v_{i} sin θ$

Instantaneous Velocity at Time $t$

The overall velocity of the projectile at any instant is the vector sum of its horizontal and vertical components at that time.

Horizontal Component ( $v_{f x}$ ): Remains unchanged. $v_{f x} = v_{i} cos θ$
Vertical Component ( $v_{f y}$ ): Changes with time. $v_{f y} = v_{i} sin θ - g t$

The magnitude of the final velocity ( $v_{f}$ ) can be found using the Pythagorean theorem: $v_{f} = v_{f x}^{2} + v_{f y}^{2} = (v_{i} cos θ)^{2} + (v_{i} sin θ - g t)^{2}$

The direction of the final velocity (the angle $ϕ$ it makes with the horizontal) is: $ϕ = tan^{- 1} (\frac{v _{f y}}{v _{f x}})$

Key Derived Quantities

Time of Flight ( $T$ )

The total time the projectile remains in the air is found by setting the vertical displacement back to zero: $T = \frac{2 v _{i} s i n θ}{g}$

Maximum Height ( $H$ )

At maximum height, the vertical velocity is zero ( $v_{f y} = 0$ ). Using $v_{f y}^{2} = v_{i y}^{2} - 2 g H$ : $H = \frac{v _{i}^{2} s i n ^{2} θ}{2 g}$

Horizontal Range ( $R$ )

The total horizontal distance covered during the flight: $R = \frac{v _{i}^{2} s i n 2 θ}{g}$

The range is maximum when $θ = 4 5^{\circ}$ (since $sin 2 θ$ is maximum at $sin 9 0^{\circ} = 1$ ).

Projectiles launched at complementary angles (e.g., $3 0^{\circ}$ and $6 0^{\circ}$ ) with the same initial speed give the same horizontal range.

Trajectory Equation

By eliminating time $t$ from the horizontal and vertical displacement equations, we obtain the equation of the trajectory: $y = x tan θ - \frac{g x ^{2}}{2 v _{i}^{2} c o s ^{2} θ}$

This is of the form $y = a x + b x^{2}$ , confirming the path is a parabola.

Summary Table

Component	Acceleration	Velocity	Displacement
Horizontal (x)	$a_{x} = 0$	$v_{x} = v_{i} cos θ$	$x = (v_{i} cos θ) t$
Vertical (y)	$a_{y} = - g$	$v_{y} = v_{i} sin θ - g t$	$y = (v_{i} sin θ) t - \frac{1}{2} g t^{2}$

Conceptual Points

Q: What force is responsible for the curved path of a projectile?

A: The constant downward force of gravity is the only force acting on an ideal projectile. It continuously changes the vertical component of the projectile's velocity, causing the parabolic trajectory.

Q: If you drop a bullet and fire another one horizontally from the same height, which one hits the ground first (ignoring air resistance)?

A: They will both hit the ground at the same time. Their horizontal motions are independent of their vertical motions. Both start with zero initial vertical velocity and accelerate downwards at the same rate ( $g$ ).

For the purpose of basic analysis, we study an ideal projectile. This model makes several simplifying assumptions:

Air resistance is negligible.
The acceleration due to gravity, g, is constant (approximately $9.8 ms^{- 2}$ ) and directed downwards.
The rotation of the Earth does not affect the motion.

Under these conditions, the trajectory of a projectile is a parabola.

Analysis of Motion: Separating the Components

The key to solving projectile motion problems is to analyze the horizontal and vertical components of motion separately. This is a practical application of Rectangular Components Of A Vector→.

Horizontal Motion

Acceleration ( $a_{x}$ ): Since we ignore air resistance, there are no horizontal forces acting on the projectile. Therefore, the horizontal acceleration is zero. $a_{x} = 0$
Velocity ( $v_{x}$ ): Because the acceleration is zero, the horizontal component of the velocity is constant throughout the flight. $v_{f x} = v_{i x} = constant$
Displacement ( $x$ ): The horizontal displacement (range) is simply the constant horizontal velocity multiplied by time. $x = v_{i x} t$

Vertical Motion

Acceleration ( $a_{y}$ ): The only force acting vertically is gravity. Therefore, the vertical acceleration is constant and directed downwards. $a_{y} = - g$
Velocity ( $v_{y}$ ): The vertical velocity changes continuously due to gravity. It can be found using the first equation of motion. $v_{f y} = v_{i y} - g t$
Displacement ( $y$ ): The vertical displacement (height) changes over time and can be found using the second equation of motion. $y = v_{i y} t - \frac{1}{2} g t^{2}$

Initial Velocity Components

If a projectile is launched with an initial velocity $v_{i}$ at an angle $θ$ above the horizontal, we must first resolve this velocity into its x and y components.

Horizontal Initial Velocity ( $v_{i x}$ ): $v_{i x} = v_{i} cos θ$
Vertical Initial Velocity ( $v_{i y}$ ): $v_{i y} = v_{i} sin θ$

Instantaneous Velocity at Time $t$

The overall velocity of the projectile at any instant is the vector sum of its horizontal and vertical components at that time.

Horizontal Component ( $v_{f x}$ ): Remains unchanged. $v_{f x} = v_{i} cos θ$
Vertical Component ( $v_{f y}$ ): Changes with time. $v_{f y} = v_{i} sin θ - g t$

The magnitude of the final velocity ( $v_{f}$ ) can be found using the Pythagorean theorem: $v_{f} = v_{f x}^{2} + v_{f y}^{2} = (v_{i} cos θ)^{2} + (v_{i} sin θ - g t)^{2}$

The direction of the final velocity (the angle $ϕ$ it makes with the horizontal) is: $ϕ = tan^{- 1} (\frac{v _{f y}}{v _{f x}})$

Key Derived Quantities

Time of Flight ( $T$ )

The total time the projectile remains in the air is found by setting the vertical displacement back to zero: $T = \frac{2 v _{i} s i n θ}{g}$

Maximum Height ( $H$ )

At maximum height, the vertical velocity is zero ( $v_{f y} = 0$ ). Using $v_{f y}^{2} = v_{i y}^{2} - 2 g H$ : $H = \frac{v _{i}^{2} s i n ^{2} θ}{2 g}$

Horizontal Range ( $R$ )

The total horizontal distance covered during the flight: $R = \frac{v _{i}^{2} s i n 2 θ}{g}$

The range is maximum when $θ = 4 5^{\circ}$ (since $sin 2 θ$ is maximum at $sin 9 0^{\circ} = 1$ ).

Projectiles launched at complementary angles (e.g., $3 0^{\circ}$ and $6 0^{\circ}$ ) with the same initial speed give the same horizontal range.

Trajectory Equation

By eliminating time $t$ from the horizontal and vertical displacement equations, we obtain the equation of the trajectory: $y = x tan θ - \frac{g x ^{2}}{2 v _{i}^{2} c o s ^{2} θ}$

This is of the form $y = a x + b x^{2}$ , confirming the path is a parabola.

Summary Table

Component	Acceleration	Velocity	Displacement
Horizontal (x)	$a_{x} = 0$	$v_{x} = v_{i} cos θ$	$x = (v_{i} cos θ) t$
Vertical (y)	$a_{y} = - g$	$v_{y} = v_{i} sin θ - g t$	$y = (v_{i} sin θ) t - \frac{1}{2} g t^{2}$

Conceptual Points

Q: What force is responsible for the curved path of a projectile?

Q: If you drop a bullet and fire another one horizontally from the same height, which one hits the ground first (ignoring air resistance)?