Understanding Motion Under Constant Acceleration

• $v_{ix} = v_i \cos \theta$
• $v_{iy} = v_i \sin \theta$

Analyzing Constant Acceleration

When acceleration is constant, the velocity changes by the same amount every second. The direction of acceleration is crucial. In projectile motion, we often ignore air resistance, meaning the only acceleration is gravity, which acts strictly in the downward direction (the negative $y$ direction). This means the horizontal component of acceleration is zero ($a_x = 0$), resulting in a constant horizontal velocity.

To solve these problems, you must identify the initial conditions. These are the values of position and velocity at the exact moment the motion begins ($t = 0$). Without knowing the initial speed ($v_i$), the launch angle ($\theta$), and the starting position ($x_i, y_i$), you cannot predict the future state of the particle.

Exam Tip: Always check if the acceleration is truly constant. If a problem mentions "air resistance," the acceleration is no longer constant because it changes based on the object's speed, and these standard kinematic equations will no longer be accurate.

Kinematic Equations and Final Velocity

By combining the initial conditions with the constant acceleration, we can derive equations to find the final velocity ($\vec{v}_f$) or the final position at any time $t$.

For the vertical direction ($y$), the velocity at any time is: $v_y = v_{iy} + a_y t$ Since gravity ($g \approx 9.8 \text{ m/s}^2$) acts downward, we substitute $a_y = -g$: $v_y = v_i \sin \theta - gt$

The vertical position is given by: $y = y_i + (v_i \sin \theta)t - \frac{1}{2}gt^2$

Worked Example: Maximum Height

Suppose a tennis ball is launched at $30 \text{ m/s}$ at an angle $\theta$. To find the maximum height ($h$), we first find the time it takes to reach the peak. At the peak, the vertical component of velocity is zero ($v_y = 0$): $0 = v_i \sin \theta - gt \implies t = \frac{v_i \sin \theta}{g}$

Substituting this time back into the position equation: $h = \frac{(v_i \sin \theta)^2}{2g}$

Notice that if you double the initial speed ($v_i$), the maximum height does not just double—it quadruples, because the velocity term is squared.

Summary of Motion Characteristics

By mastering these components, you can decompose any complex 2D motion into two simple 1D problems, making even the most difficult projectile trajectories easy to calculate.