Projectile and Planar Motion

2. Projectile Motion: A Case Study

Projectile motion is the quintessential example of 2D kinematics. It describes an object (the projectile) moving in a vertical plane under the sole influence of Earth's gravitational force. To build our model, we make two key idealizations: the acceleration due to gravity, $g$, is constant and directed downwards, and air resistance is negligible.

Initial Conditions and Setup

Consider a projectile launched from the origin $(x_0=0, y_0=0)$ with an initial speed $v_0$ at an angle $\theta$ above the horizontal. We establish a standard Cartesian coordinate system with the x-axis horizontal and the y-axis vertical.

The constant acceleration vector is purely vertical because gravity is the only force acting: $\vec{a} = 0\hat{i} - g\hat{j}$.

The initial velocity vector is resolved into its components: $\vec{v}_0 = (v_0 \cos\theta)\hat{i} + (v_0 \sin\theta)\hat{j}$.

Derivation of Kinematic Equations via Integration

We apply the integral forms of the kinematic equations separately for each component, which is possible because the motions are independent.

Step 1: Horizontal Motion (Zero Acceleration)

Because we neglect air resistance, there are no forces acting in the horizontal direction. By Newton's Second Law ($\sum F_x = ma_x$), this means the horizontal acceleration $a_x$ is zero.
Velocity: We integrate $a_x$ to find $v_x(t)$.
$v_x(t) = v_{0x} + \int_0^t a_x dt' = v_0\cos\theta + \int_0^t 0 \, dt' = v_0\cos\theta$. As expected, the horizontal velocity is constant throughout the motion.
Position: We integrate the constant horizontal velocity $v_x(t)$ to find $x(t)$.
$x(t) = x_0 + \int_0^t v_x(t') dt' = 0 + \int_0^t (v_0\cos\theta) dt' = v_0\cos\theta \int_0^t dt' = (v_0\cos\theta)t$. The horizontal position increases linearly with time.

Step 2: Vertical Motion (Constant Acceleration)

The only force acting on the projectile is gravity, so the acceleration is constant and directed downwards, $a_y=-g$.
Velocity: We integrate $a_y$ to find $v_y(t)$.
$v_y(t) = v_{0y} + \int_0^t a_y dt' = v_0\sin\theta + \int_0^t (-g) dt' = v_0\sin\theta - g\int_0^t dt' = v_0\sin\theta - gt$. The vertical velocity decreases linearly with time.
Position: We integrate $v_y(t)$ to find $y(t)$.
$y(t) = y_0 + \int_0^t v_y(t') dt' = 0 + \int_0^t (v_0\sin\theta - gt') dt' = (v_0\sin\theta)t - \frac{1}{2}gt^2$.

Step 3: The Trajectory Equation (Proof of Parabolic Path)

While the parametric equations $x(t)$ and $y(t)$ fully describe the motion, they don't immediately show the geometric shape of the path. To find the trajectory, we create an equation that relates $y$ directly to $x$ by eliminating the time variable, $t$.

Isolate Time: From the simplest equation, the horizontal position, $x = (v_0\cos\theta)t$, we solve for time:

Substitute into Vertical Equation: Now, we substitute this expression for $t$ into the vertical position equation, $y = (v_0\sin\theta)t - \frac{1}{2}gt^2$:

Simplify: The $v_0$ terms cancel in the first term, and we use the identity $\frac{\sin\theta}{\cos\theta} = \tan\theta$.

This equation is in the form $y = Ax - Bx^2$, where $A = \tan\theta$ and $B = \frac{g}{2v_0^2\cos^2\theta}$ are constants for a given launch. This is the standard form for a downward-opening parabola, providing a rigorous proof that the trajectory of an ideal projectile is parabolic.

3. Key Characteristics of Projectile Motion

Time of Flight ($T$)

The time of flight is the total duration the projectile spends in the air. Assuming it starts and lands at the same vertical level (i.e., it starts at $y=0$ and lands at $y=0$), we can set the vertical position equation to zero and solve for the time $t=T$.

We can factor out $T$: $T(v_0\sin\theta - \frac{1}{2}gT) = 0$. This equation gives two solutions for time. The first is $T=0$, which is the trivial solution representing the moment of launch. The meaningful solution comes from setting the second factor to zero:

Maximum Height ($H$)

The peak of the trajectory is the point where the projectile stops moving upward and begins to move downward. At this exact instant, its vertical component of velocity is momentarily zero. Let's call the time to reach this peak $t_p$.

Find Time to Peak: We set the vertical velocity equation $v_y(t)$ to zero: $v_y(t_p) = v_0\sin\theta - gt_p = 0$. Solving for $t_p$ gives $t_p = \frac{v_0\sin\theta}{g}$. Notice this is exactly half the total time of flight, $t_p = T/2$, which makes intuitive sense for a symmetric parabola.

Calculate Height at $t_p$: To find the maximum height $H$, we substitute this time back into the vertical position equation $y(t)$:

Horizontal Range ($R$)

The range is the total horizontal distance traveled during the time of flight $T$. Since the horizontal velocity is constant, this is a simple calculation using $d=vt$.

Calculate Range: $R = x(T) = (v_0\cos\theta)T$. Now we substitute the full expression for $T$:

Using the double-angle trigonometric identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we can write this in a more compact and elegant form:

From this equation, we can see that for a fixed initial speed $v_0$, the maximum range is achieved when the term $\sin(2\theta)$ is maximum, which is 1. This occurs when $2\theta=90^\circ$, meaning the optimal launch angle for maximum range is $\theta=45^\circ$.

4. General Curvilinear Motion

While projectile motion involves constant acceleration, an object can follow any curved path. For such general motion, it's often more intuitive to analyze the acceleration vector in a coordinate system that moves with the particle. This system is defined by unit vectors tangent ($\hat{u}_t$) and normal ($\hat{u}_n$) to the path.

• Tangential Acceleration ($\vec{a}_t$): This component is parallel to the velocity vector and is responsible for changing the particle's speed. Its magnitude is the rate of change of speed: $a_t = \frac{d|\vec{v}|}{dt}$.
• Normal (Centripetal) Acceleration ($\vec{a}_n$): This component is perpendicular to the velocity vector, pointing towards the instantaneous center of the path's curvature. It is responsible for changing the direction of the velocity vector. Its magnitude is $a_n = \frac{v^2}{\rho}$, where $\rho$ is the local radius of curvature of the path (the radius of a circle that 'kisses' the curve at that point).

5. Problem Set