Question

A batter hits a pop-up straight up in the air from a height of \(1.397 \mathrm{~m}\). The baseball rises to a height of \(7.653 \mathrm{~m}\) above the ground. Ignoring air resistance, what is the speed of the baseball when the catcher gloves it \(1.757 \mathrm{~m}\) above the ground?

Short answer

Answer: The speed of the baseball when it descends to a height of 1.757 meters is 10.90 m/s.

Step-by-step solution

Identify the given information and the required unknowns

Here, we need to find the final speed of the baseball when it reaches a height of 1.757 meters above the ground. The given parameters are: - Initial height of baseball: \(h_i = 1.397\) m - Final height of the baseball: \(h_f = 1.757\) m - Maximum height of the baseball: \(h_{max} = 7.653\) m - Acceleration due to gravity: \(a = -9.81\) m/s² The negative sign indicates acceleration in the downward direction.

Use the equation of motion to find final speed

We use the following equation of motion relating initial position, final position, initial velocity, final velocity, and acceleration: \(v_f^2 = v_i^2 + 2a(h_f - h_i)\) We do not know the initial velocity (\(v_i\)), but we can use the fact that the baseball reaches its maximum height (\(h_{max}\)) at the peak of its ascent. At this point, the vertical velocity is zero. So, we can find the initial velocity \(v_i\) when it starts descending from the maximum height and then use it to find the final velocity when it reaches the desired height.

Find the initial velocity during descent

Using the equation of motion, find \(v_i\) when the baseball descends from the maximum height (\(h_{max}\)): \(v_i^2 = 0^2 - 2a(h_{max} - h_i)\) Solving for \(v_i\), we get the magnitude of the initial velocity during descent: \(v_i = \sqrt{2a(h_{max} - h_i)}\) Plugging in the values, we find \(v_i\): \(v_i = \sqrt{2(-9.81)(7.653 - 1.397)} = 10.90\) m/s

Find the final velocity at the desired height

Now, we will use the equation of motion to find the final velocity when it reaches the height of 1.757 meters: \(v_f^2 = v_i^2 + 2a(h_f - h_i)\) Plugging in the values, we get: \(v_f^2 = (10.90)^2 + 2(-9.81)(1.757 - 7.653)\) \(v_f^2 = 118.81\) Taking the square root, we find the final velocity: \(v_f = \sqrt{118.81} = 10.90\) m/s The final speed of the baseball when the catcher gloves it 1.757 meters above the ground is 10.90 m/s.

Key concepts

Equations of Motion in Projectile Motion

In studying projectile motion, one vital tool is the equations of motion, which describe the movement of an object under the influence of forces like gravity.

Let's take the case of a pop-up hit in a baseball game. To understand and predict where and when the baseball will land, physics provides us with equations that link an object's position, velocity, acceleration, and time. These equations are a set of four, derived from the basic principles of kinematics, and in the context of projectile motion without air resistance, we're primarily interested in two of them:

• \(v = v_0 + at\) (velocity as a function of time)
• \(s = ut + \frac{1}{2}at^2\) (position as a function of time)

However, when the projectile reaches its peak, or when we're looking for final velocity at a given position, we use a third, handy equation that doesn't involve time:

• \(v_f^2 = v_i^2 + 2a(s_f - s_i)\)

This equation gives a direct relationship between velocity and position, which is very useful in problems where time is not specifically known. In the case of the baseball, by knowing the peak height and the height at which it's caught, this equation allows us to find the velocity of the baseball just before it's caught.

Acceleration Due to Gravity

An ever-present force acting on projectiles is gravity, which pulls them towards the center of the Earth. In physics problems, this force causes an acceleration due to gravity, denoted as \(g\), with a value of approximately \(9.81 \, \text{m/s}^2\) on Earth.

In our exercise, when the baseball is hit into the air, it slows down due to gravity until it stops momentarily at its highest point. From that moment, it accelerates back towards the ground. The acceleration due to gravity is constant and acts in the downward direction, which is why we assign it a negative sign when solving projectile motion problems that deal with upward movement.

This consistency of gravity allows us to make accurate predictions using the kinematic equations. To calculate the speed of the baseball when caught, we consider this acceleration in our equations of motion. By meticulously applying the value of gravity, students can ascertain the final velocities of projectiles, understanding the symmetrical nature of the paths taken by objects in freefall.

Kinematics

The branch of physics called kinematics is essentially the study of motion without considering the forces that cause it. It's all about position, velocity, and acceleration, and how these quantities are intertwined.

In projectile motion exercises, such as the baseball scenario, we use kinematic principles to relate these quantities. The initial step is always to determine what is known and what needs to be found; this will guide which mathematical relationships or kinematic equations to apply.

When you are armed with the equations of motion and a grasp of concepts like acceleration due to gravity, solving a problem becomes a matter of careful substitution and algebraic manipulation. Even though air resistance is neglected in many textbook problems, in the real world, it would certainly affect the baseball's flight, which is why more advanced physics problems might include such complexities. For now, understanding the fundamental kinematic concepts provides a solid foundation for tackling more complicated physics problems in the future.