Problem 86 - College Physics

Question

A football is kicked off the flat ground at \(25.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) relative to the ground. (a) Determine the total time it is in the air. (b) Find the angle of its velocity with respect to the ground after it has been in the air for one-fourth of this time. (c) Repeat for one-half and threefourths of the total time. (d) For each of these times, determine its speed. Comment on the speed changes as it follows its parabolic arc. Do they make sense physically?

Step-by-Step Solution

Answer

(a) 2.55 s. (b) 15.9°, (c) 0°, -15.9°. (d) Speeds: 22.53 m/s, 21.65 m/s, 22.53 m/s. Speeds are symmetric as expected.

Step 1: Determine the initial vertical and horizontal velocity components

The initial velocity of the ball can be broken into horizontal and vertical components using trigonometric functions. Let the initial velocity be \(v_0 = 25.0 \, \mathrm{m/s}\) and the launch angle \(\theta = 30^\circ\).The horizontal component \(v_{0x}\) is given by:\[ v_{0x} = v_0 \cdot \cos(\theta) \]The vertical component \(v_{0y}\) is given by:\[ v_{0y} = v_0 \cdot \sin(\theta) \]Plug in the values:\[ v_{0x} = 25.0 \cdot \cos(30^\circ) \approx 21.65 \, \mathrm{m/s} \]\[ v_{0y} = 25.0 \cdot \sin(30^\circ) \approx 12.5 \, \mathrm{m/s} \]

Step 2: Calculate the total time in the air

The time the ball spends in the air is determined by the vertical motion, assuming it returns to the same height as it was launched. Using the kinematic equation for vertical motion:\[ t_{\text{total}} = \frac{2 v_{0y}}{g} \]where \(g = 9.8 \, \mathrm{m/s^2}\) is the acceleration due to gravity.Substitute the known values:\[ t_{\text{total}} = \frac{2 \times 12.5}{9.8} \approx 2.55 \, \mathrm{s} \]

Step 3: Determine velocities at fractions of total time

We need to find the velocity components at one-fourth, one-half, and three-fourths of the total time:**One-fourth of total time:**\[ t_{1/4} = \frac{1}{4} \times 2.55 \approx 0.64 \, \mathrm{s} \]**One-half of total time:**\[ t_{1/2} = \frac{1}{2} \times 2.55 \approx 1.28 \, \mathrm{s} \]**Three-fourths of total time:**\[ t_{3/4} = \frac{3}{4} \times 2.55 \approx 1.91 \, \mathrm{s} \]The horizontal velocity \(v_x\) remains constant at \(21.65 \, \mathrm{m/s}\). The vertical velocity \(v_y\) at these times is calculated using:\[ v_y = v_{0y} - g \cdot t \]

Step 4: Calculate velocities and angles at each fraction of total time

For each fraction, calculate the vertical velocity and angle:**At one-fourth of total time:**\[ v_y = 12.5 - 9.8 \times 0.64 \approx 6.21 \, \mathrm{m/s} \]The angle \(\phi\) with respect to the ground is:\[ \tan(\phi) = \frac{v_y}{v_x} \]\[ \phi_{1/4} \approx \tan^{-1}\left(\frac{6.21}{21.65}\right) \approx 15.9^\circ \]**At one-half of total time:**\[ v_y = 12.5 - 9.8 \times 1.28 \approx 0 \, \mathrm{m/s} \]The angle \(\phi_{1/2} = 0^\circ\) since it is purely horizontal.**At three-fourths of total time:**\[ v_y = 12.5 - 9.8 \times 1.91 \approx -6.21 \, \mathrm{m/s} \]\[ \phi_{3/4} \approx \tan^{-1}\left(\frac{-6.21}{21.65}\right) \approx -15.9^\circ \]

Step 5: Calculate speed at each time

Calculate the speed at each fraction using the Pythagorean theorem:**At one-fourth of total time:**\[ v = \sqrt{v_x^2 + v_y^2} \approx \sqrt{21.65^2 + 6.21^2} \approx 22.53 \, \mathrm{m/s} \]**At one-half of total time:**\[ v = \sqrt{21.65^2 + 0^2} = 21.65 \, \mathrm{m/s} \]**At three-fourths of total time:**\[ v = \sqrt{21.65^2 + (-6.21)^2} \approx 22.53 \, \mathrm{m/s} \]

Step 6: Analyze speed changes

The speeds at one-fourth and three-fourths of the total time are equal, reflecting symmetry in the projectile's path. The horizontal component remains constant, and the vertical component's impact on speed is zero at half-time, leading to the minimum speed. These findings are consistent with the physical behavior expected in projectile motion. The symmetric path ensures similar speeds but opposite vertical directions at corresponding times.

Key Concepts

Kinematics

Kinematics is all about understanding motion without worrying about its causes. It helps us describe how objects move, using a few basic equations and concepts. One of the essential things in kinematics is separating motion into vertical and horizontal components. This distinction makes it easier to analyze the motion of objects, such as a football, in projectile motion.
When analyzing projectile motion, the vertical part is influenced by gravity, while the horizontal part moves at a constant speed. By using kinematic equations, we can find out how an object moves over time. These equations can help determine the time an object stays in the air, like how long a football remains airborne after being kicked. By understanding kinematics, we can predict the motion path and determine different factors such as speed, time, and distance throughout its journey.

Velocity Components

When dealing with projectile motion, like a football kicked at an angle, it's crucial to break down the initial velocity into horizontal and vertical components. This enables us to analyze and predict its motion accurately.
The horizontal velocity component \( v_{0x} \) remains constant throughout the flight because there are no horizontal forces acting in projectile motion, assuming air resistance is negligible. We calculate it using the cosine function, \[ v_{0x} = v_0 \cdot \cos(\theta) \].
On the other hand, the vertical velocity component \( v_{0y} \) is affected by gravity. We use the sine function to find it, \[ v_{0y} = v_0 \cdot \sin(\theta) \]. As the projectile moves, gravity constantly changes the vertical speed, eventually bringing it back to the ground with closely equal and opposite speeds but the same magnitude as it launched.

Trigonometry

Trigonometry is an essential tool in understanding projectile motion. It helps us break down the velocity into horizontal and vertical components using basic trigonometric functions such as sine and cosine. This breakdown is crucial when dealing with motions not aligned along a single axis but rather involving angles.
In our example of a football kicked at a 30-degree angle, trigonometry allows us to simplify analysis and make accurate predictions about the path of the football. By understanding how to navigate between angles and lengths, especially using \( \sin \text{ and } \cos \theta \), students can accurately calculate initial velocities, angles, and the subsequent motion paths of projectiles. This makes trigonometry a vital part of resolving the motions in different directions and predicting trajectories efficiently.

Parabolic Motion

Projectile motion typically follows a parabolic path, which means the object moves both up and sideways before coming down, forming a symmetric arc. This path results from the constant horizontal motion combined with the accelerated vertical motion due to gravity.
In this symmetrical arc, the object reaches its maximum height halfway through its flight, where its vertical velocity becomes zero before gravity starts pulling it back down. At this point, the horizontal velocity is at its maximum. The speed at one-fourth and three-fourths of the path is about the same, reflecting this symmetry. Understanding parabolic motion helps explain why the projectile returns to the ground with opposite vertical velocities at these symmetrical points in time. This concept shows the natural, predictable path that projectiles form, making studying and predicting motion manageable and straightforward.