Problem 41 - University Physics with Modern Physics

Question

Suppose you throw a 0.052 -kg ball with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(30.0^{\circ}\) above the horizontal from a building \(12.0 \mathrm{~m}\) high. a) What will be its kinetic energy when it hits the ground? b) What will be its speed when it hits the ground?

Step-by-Step Solution

Answer

Question: A ball of mass 0.052 kg is thrown off a building with an initial speed of 10.0 m/s at an angle of 30 degrees above the horizontal. The building is 12.0 m tall. Calculate a) the speed of the ball when it hits the ground and b) the kinetic energy of the ball when it hits the ground. Answer: Follow the steps outlined in the solution to calculate the initial velocity components, time of flight, horizontal and vertical velocities at impact, speed when the ball hits the ground, and the kinetic energy at impact. Remember to use the given mass, initial speed, launch angle, and height of the building in your calculations.

Step 1: Determine the initial velocity components

First, we need to find the initial horizontal and vertical components of velocity. We can do that using the given initial speed (10.0 m/s) and launch angle (30 degrees). Using trigonometry and keeping in mind that cosine is for horizontal and sine is for the vertical component: \(v_{0x} = v_{0} \cos(30^{\circ}) = 10.0 \cdot \cos(30^{\circ})\) \(v_{0y} = v_{0} \sin(30^{\circ}) = 10.0 \cdot \sin(30^{\circ})\)

Step 2: Calculate the time of flight

To find the time of flight, we can use the vertical motion equation: \(h = v_{0y}t - \frac{1}{2}gt^2\) where \(h\) is the vertical displacement (12.0 m), \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (9.81 m/s^2), and \(t\) is the time. In this case, the ball is thrown from a height, so \(h\) will be negative (-12.0 m). Rearrange the equation to solve for \(t\).

Step 3: Calculate the horizontal and vertical velocities at impact

Now, use the horizontal velocity (\(v_{0x}\)) and the time of flight (\(t\)) to determine the horizontal displacement. As the horizontal velocity remains constant throughout the motion, the horizontal velocity is the same as the initial horizontal velocity (\(v_{0x}\)). To find the vertical velocity (\(v_y\)) at impact, it's necessary to consider the gravitational acceleration during the time of flight. We can use the vertical component of the kinematic equation for the final velocity: \(v_{y} = v_{0y} - gt\)

Step 4: Calculate the speed when the ball hits the ground

With both the horizontal and vertical velocities at impact, we can find the speed of the ball by using the Pythagorean theorem: \(v = \sqrt{v_{x}^2 + v_{y}^2}\)

Step 5: Calculate the kinetic energy at impact

Finally, we can calculate the kinetic energy of the ball at the moment of impact using the formula for kinetic energy: \(K.E. = \frac{1}{2}mv^2\) where \(m\) is the mass of the ball (0.052 kg) and \(v\) is the speed at impact. After plugging in the calculated values, we'll have the answers for both parts a) and b) of the exercise.

Key Concepts

Kinematic Equations

Kinematic equations are powerful tools in physics that allow us to describe the motion of objects. They help us find unknown variables like displacement, time, velocity, and acceleration, given some initial conditions. In the context of projectile motion, these equations are especially handy. Let's look at how they apply to our example of a ball thrown from a building.

When the ball is thrown, it has both horizontal and vertical components of velocity. The horizontal motion follows a constant velocity, as there is no horizontal acceleration (assuming air resistance is negligible). The equation used here is simple:

• Horizontal displacement (\( x \)) = velocity in x direction (\( v_{0x} \)) multiplied by time (\( t \)).

For vertical motion, the ball's velocity changes due to gravity. We use the kinematic equation involving time:

• Vertical displacement (\( y \)) = \( v_{0y}t - \frac{1}{2}gt^2 \)
• Vertical velocity at any time (\( v_{y} \)) = \( v_{0y} - gt \)

These equations help us determine how long the ball will be in the air (time of flight) and the velocity components when it impacts the ground.

Kinetic Energy

Kinetic energy (K.E.) is the energy that an object possesses due to its motion. It's a crucial concept when discussing projectile motion because it helps us understand the energy transformations during the projectile's flight.

When the ball is thrown, it has an initial kinetic energy based on its speed and mass. The formula for kinetic energy is \( K.E. = \frac{1}{2}mv^2 \). Here, \( m \) represents the mass of the ball, and \( v \) is its speed. As the ball moves through the air, its kinetic energy changes.

• At the point of release, the ball's kinetic energy depends on the initial speed and mass.
• When it hits the ground, we calculate the final kinetic energy using the speed determined just before impact.

This approach allows us to see how the energy is conserved or transferred as the ball travels. The key is that the kinetic energy when the ball hits the ground reveals how the potential and kinetic energies have interchanged throughout the motion.

Trigonometry in Physics

Trigonometry in physics helps us break down vectors into components, making it easier to analyze motion, especially in projectile problems. In this exercise, we used trigonometry to split the ball’s initial velocity into horizontal and vertical components.

Given an initial speed and angle, trigonometry provides a method to find these components:

• The horizontal component (\( v_{0x} \)) is calculated as \( v_{0} \cos(\theta) \) where \( \theta \) is the angle of projection.
• Similarly, the vertical component (\( v_{0y} \)) is \( v_{0} \sin(\theta) \).

These components are crucial for analyzing the motion separately along each axis. By understanding how to apply trigonometry in these scenarios, you can solve more complex physics problems that require decomposing motion into vector components. This skill is foundational in physics, allowing us to approach problems methodically and solve them step-by-step.