Problem 117 - Fundamentals of Physics

Question

A football player punts the football so that it will have a "hang time" (time of flight) of \(4.5 \mathrm{~s}\) and land \(46 \mathrm{~m}\) away. If the ball leaves the player's foot \(150 \mathrm{~cm}\) above the ground, what must be the (a) magnitude and (b) angle (relative to the horizontal) of the ball's initial velocity?

Step-by-Step Solution

Answer

Initial velocity: 20.31 m/s at an angle of 59.24°.

Step 1: Calculate the Horizontal Velocity Component

The horizontal distance the football travels (range) is 46 m, and the time of flight is 4.5 s. Use the formula for horizontal velocity: \[ v_{x} = \frac{\text{Range}}{\text{Time}} = \frac{46}{4.5} \approx 10.22 \ \text{m/s} \]

Step 2: Determine the Vertical Velocity Component

Use the formula for vertical motion, considering the total hang time and the initial height (1.50 m above the ground). The ball must travel up and then down, totaling 4.5 s. The formula for total vertical displacement is:\[ 0 = h + v_{y} t - \frac{1}{2} g t^2 \]Where \( h = 1.5 \ \text{m} \), \( g = 9.8 \, \text{m/s}^2 \), and \( t = 4.5 \ \text{s} \). Simplifying gives:\[ 0 = 1.5 + v_{y} \cdot 4.5 - 0.5 \cdot 9.8 \cdot (4.5)^2 \]Solve for \( v_{y} \):\[ v_{y} = \frac{0.5 \cdot 9.8 \cdot (4.5)^2 - 1.5}{4.5} \ \approx 17.60 \ \text{m/s} \]

Step 3: Calculate the Magnitude of the Initial Velocity

The initial velocity \( v \) is found by combining the horizontal and vertical components using the Pythagorean theorem:\[ v = \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{(10.22)^2 + (17.60)^2} \approx 20.31 \ \text{m/s} \]

Step 4: Find the Angle of Projection

Use the tangent function to find the angle \( \theta \) of the initial velocity relative to the horizontal:\[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = \tan^{-1}\left(\frac{17.60}{10.22}\right) \approx 59.24^\circ \]

Key Concepts

Horizontal Velocity

In projectile motion, the concept of horizontal velocity is crucial as it remains constant throughout the flight, assuming no air resistance. This velocity determines how far the projectile will travel horizontally before it lands. In our exercise, the football travels a horizontal distance of 46 meters in 4.5 seconds.
To find the horizontal velocity (\( v_{x} \)), use the formula: \[ v_{x} = \frac{\text{Range}}{\text{Time}}\]Here, the range is the distance traveled horizontally, and the time is the total duration of flight. Plugging in the values:\[ v_{x} = \frac{46}{4.5} \approx 10.22 \text{ m/s} \]

• The horizontal velocity is unaffected by gravity.
• It solely depends on the initial speed and angle.

Understanding horizontal velocity helps to predict whether an object will reach its target.

Vertical Velocity

Vertical velocity refers to the speed at which the football moves upward or downward during its flight. Initially, when the ball is kicked, it has an upward motion that gradually decreases to zero at the peak, then increases downward.
Using the vertical motion formula with an initial height of 1.5 meters above the ground:\[ 0 = h + v_{y} t - \frac{1}{2} g t^2 \]Here, \( h = 1.5 \ ext{m}, g = 9.8 \ ext{m/s}^2, t = 4.5 \ ext{s} \). By rearranging and solving for \( v_{y} \):\[ v_{y} = \frac{0.5 \cdot 9.8 \cdot (4.5)^2 - 1.5}{4.5} \approx 17.60 \text{ m/s}\]

• Vertical velocity changes due to gravitational acceleration.
• Maximum height occurs when vertical velocity is zero.

The concept is key to understanding how high the ball travels.

Initial Velocity

The initial velocity of a projectile is the speed at which it is launched. It's crucial as it defines the trajectory of the football. The exercise requires finding both the magnitude and direction of this initial speed.
Combine the horizontal and vertical components with the Pythagorean theorem:\[ v = \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{(10.22)^2 + (17.60)^2} \approx 20.31 \text{ m/s} \]

• Initial velocity affects how far and high a projectile travels.
• Consists of both horizontal and vertical parts.

This calculation helps predict the likely outcome of the ball's trajectory when kicked.

Angle of Projection

The angle of projection is the initial angle at which a projectile is launched relative to the horizontal. This angle has a profound impact on the range and height of the projectile's path.
To find the angle, use the tangent function to relate horizontal and vertical components:\[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = \tan^{-1}\left(\frac{17.60}{10.22}\right) \approx 59.24^\circ \]

• Angles affect both the range and maximum height.
• 45 degrees typically maximizes horizontal distance (ignoring air resistance and initial height).

Understanding this angle helps adjust the angle to achieve a desired distance.