91Ó°ÊÓ

Kinematic equations are essential tools used to describe the motion of objects in various types of motion, including projectile motion. These equations relate the displacement, initial velocity, final velocity, acceleration, and time of an object's motion. For an object projected into the air like a baseball, kinematic equations can predict where it will be at specific times and how fast it will be moving. In this exercise, the vertical motion equation is used to correlate the vertical displacement of the baseball with time: \[ y = v_{0y} t + \frac{1}{2} a t^2 \] where:

• \( y \) is the vertical displacement, here given as 10.0 m.
• \( v_{0y} \) is the initial vertical velocity.
• \( a \) is the acceleration due to gravity, commonly taken as \(-9.8 \text{ m/s}^2\).
• \( t \) is the time at which we want to find the displacement or velocity.

Velocity components divide the velocity vector into horizontal and vertical pieces, making it easier to analyze motion in each direction separately. For a baseball hit at an angle:

• The horizontal velocity component \( v_{0x} \) = \( v_0 \cos \theta \).
• The vertical velocity component \( v_{0y} \) = \( v_0 \sin \theta \).

These components are critical because they don't affect each other; vertical motion is influenced by gravity, while horizontal motion is constant in the absence of air resistance. For the given exercise, we used these relations to find:

• Horizontal speed: 24.0 m/s.
• Initial vertical speed: 18.0 m/s.

This separation simplifies calculations, especially when implementing kinematic equations for different axes.

The quadratic formula is a mathematical solution used to solve second-order polynomials of the form \( ax^2 + bx + c = 0 \). It's particularly useful in projectile motion when solving for time because the motion's equations are quadratic in relation to time. For our baseball problem, we applied the formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where:

• \( a = 4.9 \), related to the acceleration due to gravity divided in half.
• \( b = -18.0 \), derived from the initial vertical speed.
• \( c = 10.0 \), representing the desired height minus the launch point height.

The roots found, \( t_1 = 0.67 \text{ s} \) and \( t_2 = 3.06 \text{ s} \), tell us the two times the ball reaches this particular height.

Vertical displacement in projectile motion refers to the change in position of an object along the vertical axis. It helps determine how far up or down an object has traveled relative to its starting point. The vertical displacement is directly influenced by the initial vertical velocity and gravity. In our problem, we calculated the displacement using a kinematic equation, focusing on reaching a specific height (10.0 m). The formula specifically outlines how displacement changes over time due to the interplay between the initial vertical velocity and the acceleration due to gravity. This naturally yields a path that is symmetric around the peak of the trajectory. Thus, by finding the times when this displacement is achieved, we understand more about the overall arc and the time frames involved in the projectile's motion.