Problem 17 - College Physics

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 \(\mathrm{m} / \mathrm{s}\) and at an angle of \(36.9^{\circ}\) above the horizontal. You can ignore air resistance. (a) At what two times is the base- ball at a height of 10.0 \(\mathrm{m}\) above the point at which it left the bat?(b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times you found in part (a). (c) What are the magnitude and direction of the base- ball's velocity when it returns to the level at which it left the bat?

Step-by-Step Solution

Answer

(a) At times 0.67 s and 3.06 s. (b) Velocity components: at 0.67 s (24.0 m/s, 11.4 m/s), at 3.06 s (24.0 m/s, -12.0 m/s). (c) Velocity: 30.0 m/s, direction -36.9°.

Step 1: Identify the Problem and Known Values

We are given a baseball hit with an initial velocity of 30.0 m/s at an angle of \(36.9^\circ\) from the horizontal. We need to find the times when the baseball is 10.0 m above the launch point, as well as the horizontal and vertical velocity components at these times. Finally, we want to find the velocity magnitude and direction when it returns to the initial level.

Step 2: Decompose Initial Velocity

Decompose the initial velocity into horizontal and vertical components:- Horizontal component: \( v_{0x} = v_0 \cos \theta = 30.0 \cos 36.9^\circ \)- Vertical component: \( v_{0y} = v_0 \sin \theta = 30.0 \sin 36.9^\circ \)

Step 3: Calculate Horizontal and Vertical Velocity Components

Calculate the components using trigonometric values:- \( v_{0x} = 30.0 \times 0.8 = 24.0 \text{ m/s} \)- \( v_{0y} = 30.0 \times 0.6 = 18.0 \text{ m/s} \)

Step 4: Apply Kinematic Equation for Vertical Motion

Use the vertical motion equation: \( y = v_{0y} t + \frac{1}{2} a t^2 \). Given, \( y = 10.0 \text{ m} \), \( a = -9.8 \text{ m/s}^2 \), solve for \( t \):\[ 10.0 = 18.0 t - 4.9 t^2 \]

Step 5: Solve Quadratic Equation for Time

Rearrange into standard form: \( 4.9 t^2 - 18.0 t + 10.0 = 0 \). Solve using the quadratic formula:\[ t = \frac{-(-18.0) \pm \sqrt{(-18.0)^2 - 4 \times 4.9 \times 10.0}}{2 \times 4.9} \]The solutions: \( t_1 = 0.67 \text{ s} \) and \( t_2 = 3.06 \text{ s} \)

Step 6: Calculate Velocity Components at t1 and t2

Use the vertical velocity formula: \( v_{y} = v_{0y} + at \).1. At \( t_1 = 0.67 \text{ s}\): \( v_{x1} = 24.0 \text{ m/s} \), \( v_{y1} = 18.0 + (-9.8)(0.67) = 11.4 \text{ m/s} \)2. At \( t_2 = 3.06 \text{ s}\): \( v_{x2} = 24.0 \text{ m/s} \), \( v_{y2} = 18.0 + (-9.8)(3.06) = -12.0 \text{ m/s} \)

Step 7: Calculate Velocity at Return to Initial Level

When the baseball returns to the launch level, vertical displacement \( y = 0 \). Its speed will be equal in magnitude to the initial speed due to symmetry in projectile motion but in opposite direction:- Horizontal velocity: \( v_x = 24.0 \text{ m/s} \)- Vertical velocity: \( v_y = -18.0 \text{ m/s} \)Calculate the magnitude: \( v = \sqrt{v_x^2 + v_y^2} = 30.0 \text{ m/s} \)Calculate direction: \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)\approx -36.9^\circ \)

Key Concepts

Kinematic Equations

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

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.

Quadratic Formula

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

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.