AP Physics 1: Angled Projectile Motion

Mastering the physics of a basketball's arc or a soccer ball's trajectory requires understanding angled projectile motion. This foundational concept connects the elegant mathematics of two-dimensional motion to countless real-world phenomena, from sports to engineering. Success in AP Physics 1 hinges on moving beyond simple horizontal launches to analyze objects projected at any angle, which demands a systematic approach to problem-solving.

Resolving the Initial Velocity Vector

Every angled projectile problem begins with a single, crucial step: breaking the initial velocity ($v_0$) into its perpendicular horizontal ($v_{0x}$) and vertical ($v_{0y}$) components. This resolution is essential because, in ideal projectile motion (neglecting air resistance), the horizontal and vertical motions are completely independent of each other. The horizontal velocity remains constant, while the vertical velocity changes at a constant rate due to gravity ($g=-9.8{\text{ m/s}}^2$).

You use right-triangle trigonometry on the initial velocity vector to find these components. If the launch angle ($\theta$) is measured from the horizontal, the components are:

• $v_{0x}=v_0\cos (\theta )$
• $v_{0y}=v_0\sin (\theta )$

For example, a football kicked with an initial speed of $20\text{ m/s}$ at a $37^{\circ }$ angle has components: $v_{0x}=20\cos (37^{\circ })\approx 16\text{ m/s}$ and $v_{0y}=20\sin (37^{\circ })\approx 12\text{ m/s}$. These component values become the starting points for all subsequent calculations of flight time, height, and range.

Key Kinematic Quantities: Time, Height, and Range

Once you have the components, you apply the constant-acceleration kinematic equations separately to the vertical and horizontal motions. Three key quantities define the projectile's journey:

1. Total Flight Time ($t_{total}$): The time from launch to landing is determined entirely by the vertical motion. For a projectile that lands at the same vertical height from which it was launched (a level surface), the vertical displacement is zero. Using the kinematic equation $\Delta y=v_{0y}t+\frac{1}{2}{a}_y{t}^2$ with $\Delta y=0$ and $a_y=-g$, you solve for time. The non-zero solution is: $$t_{total}=\frac{2v_{0y}}{g}=\frac{2v_0\sin (\theta )}{g}$$ This shows flight time depends only on the initial vertical velocity and gravity.

2. Maximum Height ($H$): At the peak of its trajectory, the projectile's vertical velocity instantaneously becomes zero ($v_y=0$). Using the kinematic equation $v_y^2=v_{0y}^2+2a_y\Delta y$, you solve for the vertical displacement $\Delta y$, which is the maximum height: $$H=\frac{v_{0y}^2}{2g}=\frac{(v_0\sin (\theta ){)}^2}{2g}$$ The maximum height depends on the square of the vertical component, so doubling $v_{0y}$ quadruples the peak height.

3. Horizontal Range ($R$): The range is the total horizontal distance traveled. Since horizontal motion has constant velocity, range is simply: $$R=v_{0x}\cdot t_{total}=(v_0\cos (\theta ))\cdot \left(\frac{2v_0\sin (\theta )}{g}\right)$$ This simplifies to the classic range equation: $$R=\frac{v_0^2\sin (2\theta )}{g}$$ This powerful equation reveals how range depends on the square of the initial speed and the sine of twice the launch angle.

Analysis of Launch Angle and Range Symmetry

The range equation $R=\frac{v_0^2\sin (2\theta )}{g}$ provides profound insights into the relationship between launch angle and distance.

For a given initial speed $v_0$, the range is maximized when $\sin (2\theta )$ is at its maximum value of 1. This occurs when $2\theta =90^{\circ }$, or $\theta =45^{\circ }$. Therefore, on a level surface, a $45^{\circ }$ launch angle yields the maximum possible range.

The sine function also creates range symmetry. Because $\sin (2\theta )=\sin (180^{\circ }-2\theta )=\sin (2(90^{\circ }-\theta ))$, two different launch angles that are complementary (i.e., they add to $90^{\circ }$) will yield the same horizontal range for the same $v_0$. For instance, a projectile launched at $30^{\circ }$ will travel the same horizontal distance as one launched at $60^{\circ }$, assuming a level launch and landing. However, their trajectories differ dramatically: the $60^{\circ }$ launch will have a longer flight time and a much higher maximum height than the $30^{\circ }$ launch.

Solving Complex Angled Launch Problems

Real AP and engineering problems often involve launches from or to different heights. In these cases, the simplifying condition $\Delta y=0$ no longer applies, so you must use the full kinematic equations.

Worked Example (Launch from a Height): A baseball is thrown from a cliff $50\text{ m}$ high with an initial velocity of $25\text{ m/s}$ at a $30^{\circ }$ angle above the horizontal. Find its time of flight and horizontal distance from the base of the cliff when it lands.

Step 1: Resolve Components. $v_{0x}=25\cos (30^{\circ })\approx 21.65\text{ m/s}$ $v_{0y}=25\sin (30^{\circ })=12.5\text{ m/s}$

Step 2: Solve for Time of Flight ($\Delta y=-50\text{ m}$). Use $\Delta y=v_{0y}t+\frac{1}{2}{a}_y{t}^2$. $-50=12.5t+\frac{1}{2}(-9.8)t^2$ Rearrange: $4.9t^2-12.5t-50=0$ Solve the quadratic: $t\approx 5.18\text{ s}$ (taking the positive root).

Step 3: Solve for Horizontal Range. $R=v_{0x}\cdot t=21.65\text{ m/s}\cdot 5.18\text{ s}\approx 112\text{ m}$

This process—resolve, select the correct kinematic equation for the vertical motion to find time, then use horizontal motion to find range—is the universal method for solving any angled projectile problem.

Common Pitfalls

1. Mixing Horizontal and Vertical Quantities: The most critical error is using a vertical variable (like $v_{0y}$) in a horizontal equation, or vice versa. Remember: Horizontal motion has constant velocity ($a_x=0$). Vertical motion has constant acceleration ($a_y=-g$). Keep your kinematic equations segregated by direction.

1. Misapplying the Symmetry Shortcuts: The symmetry properties and the simplified time-of-flight equation ($t_{total}=2v_{0y}/g$) apply only when the launch and landing heights are equal. If a projectile is launched from a cliff or lands in a ditch, you must use the full kinematic equations with the correct vertical displacement ($\Delta y\ne 0$).

1. Sign Errors with Gravity and Displacement: Consistently define your coordinate system. If "up" is positive, then acceleration due to gravity is $a_y=-9.8{\text{ m/s}}^2$. Vertical displacement ($\Delta y$) is positive if the final position is above the initial position, and negative if below. Mixing these signs is a frequent source of calculation mistakes.

1. Forgetting that $v_y=0$ at the Peak: At maximum height, the vertical velocity is zero, but the horizontal velocity remains constant and is not zero. The object is still moving horizontally at its initial $v_{0x}$ value. A related mistake is assuming the acceleration is zero at the peak—it is not; gravity is still acting.

Summary

• The First Step is Always Resolution: Use $v_{0x}=v_0\cos (\theta )$ and $v_{0y}=v_0\sin (\theta )$ to break the initial velocity into independent horizontal and vertical components.
• Time is Vertical, Range is Horizontal: Total flight time is governed by vertical kinematics. Once time is found, the horizontal range is calculated using constant velocity: $R=v_{0x}\cdot t_{total}$.
• The 45° Rule: On a level surface, a launch angle of $45^{\circ }$ provides the maximum horizontal range for a given initial speed.
• Range Symmetry Exists for Complementary Angles: Two launch angles that sum to $90^{\circ }$ (e.g., $30^{\circ }$ and $60^{\circ }$) produce the same range on a level surface, though their time-of-flight and maximum height differ.
• Master the General Method: For non-level launches, use the vertical kinematic equation with the correct $\Delta y$ to solve for time, then find the range. This method works universally.