Question

Two baseballs are thrown off the top of a building that is \(7.25 \mathrm{~m}\) high. Both are thrown with initial speed of 63.5 mph. Ball 1 is thrown horizontally, and ball 2 is thrown straight down. What is the difference in the speeds of the two balls when they touch the ground? (Neglect air resistance.)

Short answer

Answer: The difference in speeds of the two baseballs when they hit the ground is 9.48 m/s.

Step-by-step solution

Convert the initial speed from mph to m/s

The initial speed of both baseballs is given in mph. To convert this to m/s, we can use the conversion factor 1 mph = 0.44704 m/s. Initial speed = \(63.5 \mathrm{~mph} \times 0.44704\frac{\mathrm{m}}{\mathrm{s}} = 28.39 \mathrm{~m/s}\)

Calculate the time taken for each baseball to reach the ground

First, we need to find the time it takes for both baseballs to reach the ground. As they start from the same height and are subject only to the acceleration due to gravity (g), their vertical motion will be identical. We can use the following equation of motion: \(h = \frac{1}{2}gt^2\) where \(h\) is the building's height (7.25 m) and \(g\) is the acceleration due to gravity (9.81 m/s²). Solve for \(t\): \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 7.25}{9.81}} = 1.214\mathrm{~s}\)

Calculate the final speed of ball 1 (horizontal throw)

The horizontal motion of ball 1 does not have any acceleration (neglecting air resistance), so its horizontal component of velocity remains constant at 28.39 m/s. For the vertical motion, ball 1 is subject to the acceleration due to gravity (g). We can use the following equation of motion to find the final vertical component of its velocity: \(v_y = gt\) where \(v_y\) is the final vertical component of velocity and \(g\) is the acceleration due to gravity (9.81 m/s²). \(v_y = (9.81)(1.214) = 11.90\mathrm{~m/s}\) Since the horizontal and vertical components of velocity are perpendicular, we can use the Pythagorean theorem to find the final speed of ball 1: \(v_1 = \sqrt{v_x^2 + v_y^2} = \sqrt{(28.39)^2 + (11.90)^2} = 30.81\mathrm{~m/s}\)

Calculate the final speed of ball 2 (straight down throw)

For ball 2, the initial vertical velocity is equal to the initial speed, as it is thrown straight down, so \(v_{2i} = 28.39\mathrm{~m/s}\). To find the final vertical velocity of ball 2, we can use the following equation of motion: \(v_{2f} = v_{2i} + gt\) where \(v_{2f}\) is the final vertical velocity and \(g\) is the acceleration due to gravity (9.81 m/s²). \(v_{2f} = 28.39 + (9.81)(1.214) = 40.29\mathrm{~m/s}\) As ball 2 is thrown straight down, its final speed is equal to its final vertical velocity: \(v_2 = 40.29\mathrm{~m/s}\)

Calculate the difference in speeds

Now that we have the final speeds of both balls, we can find the difference between their speeds: Difference in speeds = \(|v_2 - v_1| = |40.29 - 30.81| = 9.48\mathrm{~m/s}\) The difference in speeds of the two baseballs when they hit the ground is 9.48 m/s.

Key concepts

Equations of Motion

In the realm of physics, and particularly when studying projectile motion, equations of motion are mathematical relationships that allow us to predict the future position, velocity, and acceleration of an object. They mainly rely on the initial conditions of the object's movement, like its initial velocity and position, and consider the effects of constant acceleration.

For an object in free fall, such as the baseballs in this exercise, the only force acting upon them is gravity (when neglecting air resistance). This induces a constant acceleration downwards, known as the acceleration due to gravity. One primary equation of motion used in the solution is
\(h = \frac{1}{2}gt^2\),
where
\(h\) is the height,
\(g\) is the gravitational acceleration, and
\(t\) is the time the object is in motion. Because the acceleration due to gravity is constant, this equation elegantly connects the dots between the distance an object has fallen and the time it has been falling.

Understanding these equations is crucial for making accurate predictions about an object's projectile path and final speed upon impact.

Kinematics

Kinematics is the branch of classical mechanics that describes the motion of points, objects, and systems of bodies without considering the forces that cause them to move. It's like the 'story' of an object's journey, detailing how the position of an object changes with time.

In projectile motion, kinematics tells us how the baseball moves horizontally and vertically through the space around it. The initial velocity of the ball (here converted to meters per second) is crucial for understanding both axes of its motion. For the horizontally thrown baseball, the horizontal velocity remains constant, and a separate kinematic equation accounts for its vertical acceleration due to gravity:
\(v_y = gt\).
The interplay of these kinematic principles allows us to calculate the final speed of the baseball by considering the two-dimensional aspects of its travel.

By analyzing the baseball's kinematics, we piece together the full motion, illustrating the influence of both the horizontal throw and the acceleration of gravity.

Acceleration Due to Gravity

The acceleration due to gravity, denoted by
\(g\),
is a constant value that represents the acceleration that Earth's gravitational field exerts on any object towards its center. On Earth, this value is approximately
\(9.81 \text{m/s}^2\),
although it can slightly vary depending on location and altitude.

This acceleration plays a pivotal role in projectile motion. For instance, both baseballs in our exercise are subjected to this acceleration, which affects their vertical motion. It's the reason why the downward speed of the ball thrown straight down increases over time and why the ball thrown horizontally starts to drop the moment it leaves the hand, eventually hitting the ground. Gravitational acceleration is an excellent example of a kind of 'invisible hand' that governs the vertical component of projectile motion, and is the key reason both balls hit the ground at the same time despite one being thrown downward.