Question

A batter hits a pop-up straight up in the air from a height of \(1.273 \mathrm{~m}\). The baseball rises to a height of \(7.777 \mathrm{~m}\) above the ground. The speed of the baseball when the catcher gloves it is \(10.73 \mathrm{~m} / \mathrm{s}\). At what height above the ground did the catcher glove the ball?

Short answer

Answer: The catcher gloves the baseball at approximately 6.272 m above the ground.

Step-by-step solution

Identify the variables

We are given the following variables: - Initial height of the ball from the ground: \(h_i = 1.273 \mathrm{~m}\) - Maximum height of the ball from the ground: \(h_m = 7.777 \mathrm{~m}\) - Speed of the baseball when the catcher gloves it: \(v_c = 10.73 \mathrm{~m}/\mathrm{s}\) - We need to find the height of the ball when the catcher gloves it: \(h_c\)

Find the initial and maximum gravitational potential energy

Let's find the gravitational potential energy (\(U_i\) and \(U_m\)) for the initial and the maximum height, respectively. The gravitational potential energy is given by \(U = mgh\) where \(m\) is the mass of the ball, \(g = 9.81 \mathrm{~m}/\mathrm{s}^2\) is the acceleration due to gravity, and \(h\) is the height above the ground. Note that the mass of the baseball will not change, and it will cancel out when comparing energy at different heights, so we do not need to know its value. We can write the potential energy at both heights as: \(U_i = mgh_i\) \(U_m = mgh_m\)

Calculate the maximum kinetic energy

When the baseball reaches its maximum height, its vertical speed becomes zero momentarily. Thus, using the conservation of mechanical energy principle, all the initial energy has been converted into potential energy and we can write: \(K_i + U_i = U_m\) where \(K_i\) is the initial kinetic energy of the ball. Since \(K_i = \frac{1}{2}mv_i^2\), we can rewrite the energy conservation equation as: \(\frac{1}{2}mv_i^2 + mgh_i = mgh_m\) Now, we can solve for the initial speed \(v_i\): \(v_i^2 = 2g(h_m - h_i)\)

Calculate the height at which the catcher gloves the ball

When the catcher gloves the ball, the speed of the ball is given, so we can determine its kinetic and potential energy at that moment using the conservation of mechanical energy principle: \(K_c + U_c = K_i + U_i\) where \(K_c = \frac{1}{2}mv_c^2\) \(U_c = mgh_c\) We are looking for \(h_c\). Plugging in the expression for \(v_i^2\) from step 3, we can rewrite the energy conservation equation as: \(\frac{1}{2}mv_c^2 + mgh_c = \frac{1}{2}m(2g(h_m - h_i)) + mgh_i\) Now, cancel out the mass on both sides and solve for \(h_c\): \(h_c = \frac{1}{2}(v_c^2 - 2g(h_m - h_i)) + h_i\)

Plug in the given values and solve for \(h_c\)

Now that we have a formula for \(h_c\), we can plug in the given values and solve the equation: \(h_c = \frac{1}{2}((10.73 \mathrm{~m}/\mathrm{s})^2 - 2(9.81 \mathrm{~m}/\mathrm{s}^2)(7.777 \mathrm{~m} - 1.273 \mathrm{~m})) + 1.273 \mathrm{~m}\) \(h_c \approx 6.272 \mathrm{~m}\) So, the catcher gloves the ball at a height of approximately \(6.272 \mathrm{~m}\) above the ground.

Key concepts

Mechanical Energy Conservation

Understanding the concept of mechanical energy conservation is like having a secret key to unlock many mysteries in physics. It's a principle that tells us that in a closed system, where no external forces like friction or air resistance are doing work, the total mechanical energy remains constant.

Imagine mechanical energy as a currency that can be traded between two forms: potential energy, which you can think of as 'stored' energy, and kinetic energy, the energy of motion. Despite this trade-off, the total 'wealth'—the mechanical energy—doesn't change. For our case, when a baseball is hit into the air, the energy it got from the bat is conserved, transforming between gravitational potential energy and kinetic energy as the ball rises and falls. This principle allows us to set up equations that relate the height of the ball and its speed at different points in its journey through the air, like in the exercise we have at hand.

Let's simplify this with a scenario: If a child had 10 marbles and divided them between two boxes, no matter how they split them up, they still have 10 marbles in total. Similarly, the baseball's initial kinetic energy and potential energy at the batter's level, when combined, will equal its potential energy at the peak of its flight where its speed is zero.

Kinetic Energy

Now, let's zoom in on one half of our energy trading system—kinetic energy. Kinetic energy is the energy an object possesses due to its motion. It’s as if the object is carrying energy around because it’s moving. The faster the object moves, the more kinetic energy it has—it's like when you run; the faster you go, the more energy you're using.

The formula to calculate kinetic energy (\( K \)) is \( K = \frac{1}{2}mv^2 \), where ‘m’ stands for the mass of the object and ‘v’ represents its velocity. In our baseball example, as the ball is caught, it has kinetic energy because of its velocity. Even though we don’t know the mass of the ball, we don’t need it. That's because when we compare kinetic energy at different points, the mass cancels out, just like the price tag gets removed when we exchange presents. We use the final velocity of the ball, given at the moment the catcher gloves it, to determine its kinetic energy at that exact instant.

Acceleration Due to Gravity

Diving into the concept of 'acceleration due to gravity' adds an exciting twist to our story of the baseball's flight. It is the force that pulls everything to the ground and is why the baseball didn't just keep rising forever. On Earth, this acceleration is approximately \( 9.81 \text{ m/s}^2 \) and is symbolized by the letter ‘g’. It's the universal rate at which objects accelerate towards the ground, sans resistance.

Whether it’s a feather or a hammer, if dropped from the same height and in the absence of air resistance, both would hit the ground at the same time. This acceleration is a fundamental part of the equations we use to calculate both kinetic and potential energy. For instance, the potential energy of the baseball at any height is given by the product of its mass, the acceleration due to gravity, and its height above the ground, as shown in \( U = mgh \). Despite its constant value, 'g' plays a dynamic role in the energetic dance of our baseball. It's a silent partner in the exchange between kinetic and potential energy, contributing to the captivating tale of the baseball's rise and fall.