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Chapter 5: Problem 25

Football Kicker A football kicker can give the ball an initial speed of \(25 \mathrm{~m} / \mathrm{s}\). Within what two elevation angles must he kick the ball to score a field goal from a point \(50 \mathrm{~m}\) in front of goalposts whose horizontal bar is \(3.44 \mathrm{~m}\) above the ground? (If you want to work this out algebraically, use \(\sin ^{2} \theta+\cos ^{2} \theta=1\) to get a relation between \(\tan ^{2} \theta\) and \(1 / \cos ^{2} \theta\), substitute, and then solve the resulting quadratic equation.)

Short Answer

Expert verified
The kicker should use elevation angles of approximately 18.57掳 or 71.43掳.

Step by step solution

01

Identify Key Variables and Equations

Given:- Initial speed: 25 m/s- Distance to goalposts: 50 m- Height of goalposts bar: 3.44 mKey equations:- Horizontal distance: \[ x = v_0 \times \frac{t}{\tan \theta} \] - Vertical distance: \[ y = v_0 t \tan \theta - \frac{1}{2} g t^2 \]
02

Express Time in Terms of Distance and Angle

Rearrange the horizontal distance formula to solve for time, t:\[ t = \frac{x}{v_0 \times \frac{1}{\tan \theta}} = \frac{x \tan \theta}{v_0} \]Substitute the given values:\[ t = \frac{50 \tan \theta}{25} = 2 \tan \theta \]
03

Substitute Time into Vertical Distance Equation

Substitute the expression for t into the vertical distance formula:\[ y = v_0 \left( 2 \tan \theta \right) \tan \theta - \frac{1}{2} g \left( 2 \tan \theta \right)^2 \]Simplify the equation:\[ 3.44 = 50 \tan^2 \theta - 19.6 \tan^2 \theta \]Combine and simplify further:\[ 3.44 = 30.4 \tan^2 \theta \]
04

Shift Terms and Solve Quadratic Equation

Rearrange the equation to solve for \( \tan^2 \theta \) :\[ \tan^2 \theta = \frac{3.44}{30.4} \]\[ \tan^2 \theta = 0.113 \]
05

Calculate and Determine Elevation Angles

Take the square root to find \( \tan \theta \) :\[ \tan \theta = \sqrt{0.113} \approx 0.336 \]Use the arctangent function to find angles:\[ \theta_1 = \arctan(0.336) \approx 18.57^\textdegree \]For the second angle, use the complementary angle: \( 90^\textdegree - \theta \) :\[ \theta_2 = 90^\textdegree - 18.57^\textdegree = 71.43^\textdegree \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Velocity
When dealing with projectile motion, the initial velocity of the projectile is a key factor. It's represented by the symbol \( v_0 \). The initial velocity is the speed at which the projectile is launched. In our football kicker problem, the ball is kicked with an initial speed of 25 m/s. This velocity can be broken down into horizontal and vertical components, depending on the angle at which the ball is kicked.
Elevation Angle
The elevation angle, often denoted as \( \theta \), is the angle at which a projectile is launched relative to the horizontal. This angle affects both how far and how high the projectile will go. In the problem, we need to find two distinct elevation angles that allow the football to reach the goalpost 50 meters away and 3.44 meters high. These angles can be determined through trigonometric relationships and by solving related quadratic equations.
Quadratic Equation
A quadratic equation is a second-order polynomial equation in a single variable with the form \(ax^2 + bx + c = 0 \). In our problem, the relationship \( 3.44 = 30.4 \tan^2 \theta \) simplifies to a quadratic form \( \tan^2 \theta = 0.113 \). Solving this equation gives us the value of \( \tan^2 \theta \), which we then use to find the angles.
Horizontal Distance
The horizontal distance covered by a projectile (denoted as \( x \)) is influenced by the initial velocity and the elevation angle. The formula \( x = v_0 \times \frac{t}{\tan \theta} \) helps us relate these factors. By manipulating this equation, we can find how much time (\( t \)) it takes for the football to reach the goalpost, given the distance of 50 meters.
Vertical Distance
Vertical distance (denoted as \( y \)) is the height at which the projectile reaches at a particular time \( t \). The formula \( y = v_0 t \tan \theta - \frac{1}{2} g t^2 \) combines the initial velocity, angle, and the gravitational pull (denoted by \( g \), approx 9.8 \( m/s^2 \)) to determine how high the ball will travel. By substituting the time \( t = 2 \tan \theta \) into this formula, we find the vertical height 3.44 meters, which leads back to solving the quadratic equation to find the required angles.

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Most popular questions from this chapter

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