Question

A baseball is thrown from the roof of a 22.0-m-tall building with an initial velocity of magnitude 12.0 m/s and directed at an angle of 53.1\(^\circ\) above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of 53.1\(^\circ\) \(below\) the horizontal? (c) If the effects of air resistance are included, will part (a) or (b) give the higher speed?

Short answer

(a) 23.93 m/s; (b) 23.93 m/s; (c) Part (b) would likely have a higher speed with air resistance.

Step-by-step solution

Understand the problem

We need to use energy conservation to find the speed of the ball just before it strikes the ground. The ball is thrown from a height with an initial speed at an angle, and we must consider both vertical and horizontal components. Two scenarios are presented: (a) angle is above the horizontal and (b) angle is below the horizontal.

Set up energy conservation equation

For energy conservation, the initial mechanical energy (kinetic + potential energy) is equal to the final kinetic energy. The equation is: \[ \text{Initial Energy} = \text{Final Energy} \]\[ \frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 \]Here, \(m\) is the mass, \(v_i\) is the initial speed, \(v_f\) is the final speed, \(g\) is the acceleration due to gravity (9.8 m/s extsuperscript{2}), and \(h\) is the height.

Calculate components of initial velocity

Using trigonometry, find the horizontal (\(v_{ix}\)) and vertical (\(v_{iy}\)) components of the initial velocity:\[ v_{ix} = v_i \cos(\theta) \quad \text{and} \quad v_{iy} = v_i \sin(\theta) \]For \(\theta = 53.1^\circ\), above the horizontal:\[ v_{ix} = 12.0 \cos(53.1^\circ) \approx 7.2 \text{ m/s} \]\[ v_{iy} = 12.0 \sin(53.1^\circ) \approx 9.6 \text{ m/s} \]

Apply energy conservation (part a)

Substitute known values for initial velocity above the horizontal into the energy equation:\[ \frac{1}{2}m(v_{ix}^2 + v_{iy}^2) + mgh = \frac{1}{2}mv_f^2 \]Cancel \(m\) from each term:\[ \frac{1}{2}(7.2^2 + 9.6^2) + 9.8 \times 22.0 = \frac{1}{2}v_f^2 \]Solve for \(v_f\).

Simplify and solve for final speed (part a)

Calculate the initial energy:\[ \frac{1}{2}(51.84 + 92.16) + 215.6 = \frac{1}{2}v_f^2 \]\[ 72 + 215.6 = \frac{1}{2}v_f^2 \]\[ 287.6 = \frac{1}{2}v_f^2 \]Solve for \(v_f\):\[ v_f = \sqrt{2 \times 287.6} \approx 23.93 \text{ m/s} \]

Evaluate part b

For the ball thrown \(below\) the horizontal, only the vertical component of initial speed changes sign, \(v_{iy} = -9.6\) m/s. Substitute into the energy equation and solve:\[ \frac{1}{2}(7.2^2 + (-9.6)^2) + 9.8 \times 22.0 = \frac{1}{2}v_f^2 \]Notice the initial energy remains the same since the square turns negative \(v_{iy}\) positive:\[ v_f = \sqrt{2 \times 287.6} \approx 23.93 \text{ m/s} \]Thus, the result is identical to part (a).

Analyze air resistance impact (part c)

Air resistance would decrease the speed of the ball as it travels through air. However, the final speed will typically be less impacted when traveling at a steeper angle (below the horizontal), due to a shorter time and possibly less air resistance effect. Therefore, part (b) would likely have a slightly higher final speed than part (a) if air resistance were considered.

Key concepts

Projectile Motion

Projectile motion describes the path taken by an object that is launched into the air and subject to gravity. When we talk about a projectile, like a baseball, it follows a curved path known as a parabola. This trajectory is determined by its initial speed and the angle at which it is thrown.

In this exercise, the baseball is thrown from a rooftop at a specified angle either above or below the horizontal. The initial velocity has to be broken down into two components:

• Horizontal component (\( v_{ix} \)): This doesn't change because gravity only affects the vertical component.
• Vertical component (\( v_{iy} \)): This is affected by gravity, causing the projectile to slow down as it rises and speed up as it falls towards the ground.

Understanding these components is key to analyzing how the projectile moves over time. Using trigonometry, the initial velocity can be split into these two components, and thus we can fully describe the projectile's motion.

Kinetic Energy

Kinetic energy is the energy an object has due to its motion. It’s calculated by the equation \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object.

In this problem, the initial kinetic energy comes from the baseball being thrown with a known initial velocity. As the ball moves through its trajectory, its speed changes, so its kinetic energy also changes. When it reaches its maximum height, the vertical component of the velocity becomes zero, reducing the kinetic energy. However, as it falls back down, the speed increases, and so does its kinetic energy.

One key point about kinetic energy and projectile motion is that while the speed (and thus kinetic energy) changes throughout the motion, the total mechanical energy of the system is conserved, assuming no air resistance. This means the energy shifts between kinetic and potential forms but remains constant overall.

Potential Energy

Potential energy is the stored energy of an object due to its position in a force field, such as gravity. In our scenario, the potential energy is determined by the equation \( U = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the height above the ground.

Initially, when the baseball is on the roof, it has a specific amount of potential energy owing to its height above the ground level. As it is thrown into the air, its potential energy increases when ascending until it reaches its peak, and then decreases as it descends. At the highest point, potential energy is at a maximum, while kinetic energy is minimized. Conversely, just before the projectile hits the ground, its potential energy is minimized, and kinetic energy is maximized.

The conservation of mechanical energy principle helps us understand that even as potential and kinetic energies trade places throughout the motion, their sum remains constant, assuming no external forces (like air resistance) act on the projectile. This is crucial in predicting the motion and behavior of the projectile accurately.