Question

During one of the games, you were asked to punt for your football team. You kicked the ball at an angle of \(35.0^{\circ}\) with a velocity of \(25.0 \mathrm{~m} / \mathrm{s}\). If your punt goes straight down the field, determine the average speed at which the running back of the opposing team standing at \(70.0 \mathrm{~m}\) from you must run to catch the ball at the same height as you released it. Assume that the running back starts running as the ball leaves your foot and that the air resistance is negligible.

Short answer

Answer: To find the average speed the running back must run to catch the football, follow the steps: 1. Calculate the vertical component of the initial velocity: \(v_{0y} = 25.0\,\text{m/s} \times \sin(35.0^{\circ})\). 2. Find the time taken for the ball to reach the same height as it was released: \(t = \frac{2v_{0y}}{g}\). 3. Calculate the horizontal component of the initial velocity: \(v_{0x} = 25.0\,\text{m/s} \times \cos(35.0^{\circ})\). 4. Find the time taken for the ball to travel horizontally: \(t_{x} = \frac{x}{v_{0x}}\). 5. Calculate the average speed of the running back: \(v_{rb} = \frac{x}{t_{x}}\). Using these steps, you will be able to determine the required average speed of the running back to catch the ball at the same height it was released.

Step-by-step solution

Calculate the vertical component of the initial velocity

We have the initial velocity of the ball (\(v_{0} = 25.0\,\text{m/s}\)) and the angle at which it is kicked (\(\alpha = 35.0^{\circ}\)). Using trigonometry, we find the vertical component of the initial velocity (\(v_{0y}\)): \(v_{0y} = v_{0} \times \sin\alpha = 25.0\,\text{m/s} \times \sin(35.0^{\circ})\).

Find the time taken for the ball to reach the same height as it was released

Since there is no air resistance, the only force acting on the ball is gravity. The time taken for the ball to reach the maximum height and fall back to the initial height can be found using the following equation: \(v_{y} = v_{0y} - gt\), where \(v_{y}\) is the final vertical velocity, \(g\) is the acceleration due to gravity (\(9.81\,\text{m/s}^2\)), and \(t\) is the time taken. Since the ball will be falling back, the final and initial vertical velocities will have the same magnitude, but opposite directions, so \(v_{y} = -v_{0y}\). Then, we rearrange the equation to find \(t\): \(t = \frac{2v_{0y}}{g}\).

Calculate the horizontal component of the initial velocity

Using trigonometry, we find the horizontal component of the initial velocity (\(v_{0x}\)): \(v_{0x} = v_{0} \times \cos\alpha = 25.0\,\text{m/s} \times \cos(35.0^{\circ})\).

Find the time taken for the ball to travel horizontally

The horizontal distance travelled by the ball is given by the equation \(x = v_{0x}t_{x}\), where \(x\) is the horizontal distance and \(t_{x}\) is the time taken for the ball to travel horizontally. We have calculated the horizontal component of the initial velocity (\(v_{0x}\)) in Step 3, and the horizontal distance (\(x = 70.0\,\text{m}\)) is given. Rearrange the equation to find \(t_{x}\): \(t_{x} = \frac{x}{v_{0x}}\).

Calculate the average speed of the running back

The time taken for the ball to travel horizontally (\(t_{x}\)) is the time available for the running back to cover the distance (\(x\)). The average speed (\(v_{rb}\)) of the running back can be found using the equation: \(v_{rb} = \frac{x}{t_{x}}\). Following these steps, we will be able to determine the required average speed of the running back to catch the ball at the same height it was released.

Key concepts

Kinematics in Two Dimensions

In physics, kinematics is the study of motion without considering the forces that cause this motion. When it comes to projectile motion, like a football being kicked, we are dealing with kinematics in two dimensions - usually the vertical (up and down) and horizontal (side to side) motions. These two aspects of the motion are independent of each other. That is, the horizontal velocity has no effect on the vertical velocity and vice versa.

For a projectile, the horizontal motion occurs at a constant velocity, assuming no air resistance, as in our football example. This is a core part of uniformly accelerated motion, where the only acceleration is due to gravity, acting in the vertical direction. So while the horizontal speed remains unchanged, the vertical speed changes over time. We apply trigonometry in physics to resolve this two-dimensional motion into two one-dimensional motions - the horizontal and vertical components - which helps us analyze the projectile's path.

Trigonometry in Physics

Using Sine and Cosine to Find Velocity Components

In projectile motion problems, we often know the angle at which an object is launched and its initial velocity. Trigonometry comes into play to break down this velocity into two perpendicular components: the horizontal (along the x-axis) and the vertical (along the y-axis). For example, if a football is kicked at a certain angle with a certain velocity, we use the sine function to calculate the vertical component of the velocity by multiplying the magnitude of the velocity by the sine of the angle. Conversely, we use the cosine function to determine the horizontal component in the same manner.

By doing this, we are essentially using trigonometry to project the magnitude of the velocity onto two axes, giving us a clear picture of its individual contributions to overall motion. These components are crucial in predicting where the football will land and how a player must move to catch it.

Uniformly Accelerated Motion

A Closer Look at Gravity's Role

Uniformly accelerated motion refers to motion where an object is subject to a constant acceleration. In our football example, this constant acceleration is the acceleration due to gravity, which near the Earth's surface is approximately 9.81 meters per second squared. This acceleration affects only the vertical component of the projectile's motion.

The consequence of this constant acceleration is that any object thrown or kicked upwards against gravity will decelerate going up at a constant rate, come to a momentary stop at its peak, and then accelerate back down—again, at the same constant rate. By knowing the initial vertical component of velocity, we can calculate the time duration for the ball to rise and fall back down to its original height. This duration is a key factor in determining how fast and how far the ball will go, enabling us to solve for the other variables in the problem—like the running back’s required speed to catch the football.