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Chapter 3: Problem 79

A placekicker is about to kick a field goal. The ball is \(26.9 \mathrm{m}\) from the goalpost. The ball is kicked with an initial velocity of \(19.8 \mathrm{m} / \mathrm{s}\) at an angle \(\theta\) above the ground. Between what two angles, \(\theta_{1}\) and \(\theta_{2},\) will the ball clear the 2.74 -m-high crossbar? (Hint: The following trigonometric identities may be useful: \(\sec \theta=1 /(\cos \theta)\) and \(\left.\sec ^{2} \theta=1+\tan ^{2} \theta .\right)\)

Short Answer

Expert verified
The angles range between \(\theta_1\) and \(\theta_2\) where both allow the ball to clear the crossbar.

Step by step solution

01

Understand the Problem

We need to determine the angles at which the ball, kicked with an initial speed of \(19.8 \text{ m/s}\), will clear a crossbar 2.74 m high, located 26.9 m away.
02

Set Up the Projectile Motion Equation

We use the projectile motion equations. For the horizontal motion, \(x = v_{0} \cos(\theta) \cdot t\), and for the vertical motion, \(y = v_{0} \sin(\theta) \cdot t - \frac{1}{2}gt^2\), where \(g = 9.8 \, \text{m/s}^2\).
03

Solve for Time t

Given that \(x = 26.9\, \text{m}\), solve for \(t\) using \(t = \frac{x}{v_{0} \cos(\theta)}\). Substitute \(x = 26.9\, \text{m}\) and \(v_{0} = 19.8\, \text{m/s}\).
04

Substitute Time t into Vertical Equation

Substitute \(t = \frac{26.9}{19.8 \cos(\theta)}\) into \(y = v_{0} \sin(\theta) \cdot t - \frac{1}{2}gt^2\) and set \(y = 2.74\, \text{m}\).
05

Simplify the Equation

After substitution, you get \(2.74 = 19.8 \sin(\theta) \left(\frac{26.9}{19.8 \cos(\theta)}\right) - \frac{1}{2} \cdot 9.8 \left(\frac{26.9}{19.8 \cos(\theta)}\right)^2\). Simplify this to a quadratic in \(\tan(\theta)\).
06

Quadratic in \(\tan\theta\)

From the simplification, arrange the equation into the form \(a\tan^2\theta + b\tan\theta + c = 0\). Solve this quadratic equation using the quadratic formula \(\tan\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the values for \(\theta\).
07

Calculate the Angles

Using the calculated \(\tan\theta\) values, find \(\theta_1\) and \(\theta_2\) using \(\theta = \arctan(\tan\theta)\). Ensure to find two valid angles.

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Key Concepts

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

Trigonometric Identities
When working with projectile motion, trigonometric identities become very useful. They help us manipulate and solve equations related to angles and trajectories. A key identity is
  • \(\sec \theta =\frac{1}{\cos \theta}\)
  • \(\sec ^{2} \theta=1+\tan ^{2} \theta\)
These identities allow us to express trigonometric functions differently, aiding in simplification and solution of equations.
In our exercise, these identities help relate angle measurements to the horizontal and vertical components of motion. Understanding these relationships simplifies finding the angles required for the ball to clear the crossbar.
Projectile Equations
The motion of a projectile, like our football, is governed by key equations that describe its horizontal and vertical components. These are:
  • Horizontal motion: \( x = v_0 \cos(\theta) \cdot t \)
  • Vertical motion: \( y = v_0 \sin(\theta) \cdot t - \frac{1}{2}gt^2 \)
Here, \(v_0\) is the initial velocity, \(\theta\) is the launch angle, \(g\) is the acceleration due to gravity \((9.8 \text{ m/s}^2\)), and \(t\) is time.
This separation into components helps in solving for conditions like distance and height.
In our problem, these equations form the foundation to determine whether the kick will be successful.
Quadratic Equations
Quadratic equations come into play when simplifying motion equations involving angles. After substitution, the motion equations often turn into a quadratic form: \[ a \tan^2(\theta) + b \tan(\theta) + c = 0 \]
To solve this, we use the quadratic formula: \[ \tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This yields the values for \(\tan \theta\) that let us calculate the necessary angles \(\theta_1\) and \(\theta_2\).
These angles ensure the projectile meets its target criteria鈥攊n this case, clearing the crossbar.
Kinematics
Kinematics is the branch of mechanics describing motion without considering the forces causing it. In projectile motion, kinematics helps in understanding how objects move through space.
The focus is on key variables like velocity, displacement, time, and acceleration.
  • Displacement: The distance and direction of an object's change in position from the starting point.
  • Velocity: Speed with direction; for instance, our ball moves at \(19.8 \text{ m/s}\) at angle \(\theta\).
Using kinematic concepts with the projectile equations helps predict how the ball travels over time and space, making it possible to determine the launch conditions needed for successful projectile motion.

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Most popular questions from this chapter

A hot-air balloon is rising straight up with a speed of \(3.0 \mathrm{m} / \mathrm{s}\). A ballast bag is released from rest relative to the balloon at \(9.5 \mathrm{m}\) above the ground. How much time elapses before the ballast bag hits the ground?

A spacecraft is traveling with a velocity of \(v_{0 x}=5480 \mathrm{m} / \mathrm{s}\) along the \(+x\) direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the \(+x\) direction of \(a_{x}=1.20 \mathrm{m} / \mathrm{s}^{2}\), while the other gives it an acceleration in the \(+y\) direction of \(a_{y}=8.40 \mathrm{m} / \mathrm{s}^{2}\) At the end of the firing, find (a) \(v_{x}\) and (b) \(v_{y^{*}}\)

In a marathon race Chad is out in front, running due north at a speed of \(4.00 \mathrm{m} / \mathrm{s} .\) John is \(95 \mathrm{m}\) behind him, running due north at a speed of \(4.50 \mathrm{m} / \mathrm{s} .\) How long does it take for John to pass Chad?

A golfer imparts a speed of \(30.3 \mathrm{m} / \mathrm{s}\) to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest hole in one that the golfer can make, if the ball does not roll when it hits the green?

On a pleasure cruise a boat is traveling relative to the water at a speed of \(5.0 \mathrm{m} / \mathrm{s}\) due south. Relative to the boat, a passenger walks toward the back of the boat at a speed of \(1.5 \mathrm{m} / \mathrm{s} .\) (a) What are the magnitude and direction of the passenger's velocity relative to the water? (b) How long does it take for the passenger to walk a distance of \(27 \mathrm{m}\) on the boat? (c) How long does it take for the passenger to cover a distance of \(27 \mathrm{m}\) on the water?
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