Introduction to Projectile Motion in Physics

To solve these problems, we use a fundamental principle: we treat the motion as two independent components occurring simultaneously:

1. Horizontal Motion ($x$-direction): This is modeled as a constant velocity model. Because gravity acts vertically, there is no horizontal acceleration ($a_x = 0$). The position is given by: $x = (v_i \cos \alpha_i)t$
2. Vertical Motion ($y$-direction): This is modeled as a constant acceleration model. The object experiences a constant downward acceleration due to gravity ($a_y = -g$). The vertical velocity is: $v_y = (v_i \sin \alpha_i) - gt$

Exam Tip: When solving projectile problems, always resolve your initial velocity vector into its $x$ and $y$ components immediately. Remember that while the horizontal and vertical motions are independent, they are linked by a single variable: time.

Symmetric Trajectories

A symmetric trajectory occurs when a projectile is launched from and lands at the same vertical height. In these cases, the path is a perfect parabola, and we can derive specific formulas for its key characteristics.

Maximum Height ($h$)

The maximum height is the peak of the trajectory. At this exact point, the vertical velocity is momentarily zero, though the object is still moving horizontally. The height is calculated as: $h = \frac{v_i^2 \sin^2 \theta_i}{2g}$

Range ($R$)

The range is the total horizontal distance traveled by the projectile. For symmetric motion, it is expressed as: $R = \frac{v_i^2 \sin 2\theta_i}{g}$

Key Properties of Symmetric Motion

• Maximum Range: The greatest horizontal distance is achieved when the launch angle $\theta_i$ is exactly $45^\circ$.
• Complementary Angles: If you launch two projectiles at angles that add up to $90^\circ$ (such as $30^\circ$ and $60^\circ$), they will land at the same range. However, they will not have the same flight time or maximum height.

Critical Concept: A common misconception is that acceleration is zero at the highest point of a trajectory. This is false. While the vertical velocity is zero at the peak, gravity ($g$) continues to act on the object. If acceleration were zero, the object would stop moving vertically and simply float at that height.

Non-Symmetric Projectile Motion

In many real-world scenarios, the motion is not symmetric. This happens when a projectile is launched from an initial height (like a stone thrown from a building) and lands at a different height.

In non-symmetric projectile motion, you cannot use the simplified range or height formulas mentioned above. Instead, you must rely on the fundamental kinematic equations for constant acceleration.

Worked Example: The Falling Stone

Suppose a stone is thrown from a building with an initial velocity of $20\text{ m/s}$ at an angle of $30^\circ$ above the horizontal. If the building is $45\text{ m}$ tall, how long does it take to hit the ground?

1. Resolve the initial velocity components: $v_{xi} = 20 \cos 30^\circ = 17.3\text{ m/s}$ $v_{yi} = 20 \sin 30^\circ = 10.0\text{ m/s}$

2. Use the vertical displacement equation: Let the ground be $y_f = 0$ and the starting height be $y_i = 45\text{ m}$. $\Delta y = y_f - y_i = -45\text{ m}$ Using $\Delta y = v_{yi}t - \frac{1}{2}gt^2$: $-45 = 10t - 4.9t^2$ Rearranging into a quadratic equation: $4.9t^2 - 10t - 45 = 0$ Solving for $t$ gives approximately $4.22\text{ s}$.

3. Find the horizontal distance (Range): $x = v_{xi}t = (17.3\text{ m/s})(4.22\text{ s}) = 73.0\text{ m}$

By breaking the problem into its horizontal and vertical parts, even complex, non-symmetric paths become manageable through systematic calculation.