/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A football kicker can give the b... [FREE SOLUTION] | 91Ó°ÊÓ

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A football kicker can give the ball an initial speed of \(25 \mathrm{~m} / \mathrm{s}\). What are the (a) least and (b) greatest elevation angles at which he can 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?

Short Answer

Expert verified
\(\theta_{min} ≈ 23.8°\) and \(\theta_{max} ≈ 66.2°\).

Step by step solution

01

Identify the Variables and Known Values

Let's identify what we know from the problem. The initial speed \(v_0\) of the football is \(25\, \mathrm{m/s}\), the horizontal distance \(R\) to the goalposts is \(50\, \mathrm{m}\), and the height \(h\) of the goal bar is \(3.44\, \mathrm{m}\). We are looking for the angles \(\theta\) that will allow the football to reach exactly this height at this distance.
02

Use the Projectile Motion Equation for Range

The range equation in projectile motion is given by \(R = \frac{v_0^2 \sin(2\theta)}{g}\), where \(g = 9.81\, \mathrm{m/s^2}\) is the acceleration due to gravity. Solving for \(\theta\), we have:\[ \sin(2\theta) = \frac{R \cdot g}{v_0^2} \]Plugging in the values:\[ \sin(2\theta) = \frac{50 \times 9.81}{25^2} = 0.7848 \]
03

Solve for the Angle 2\theta

Solving for \(2\theta\) using the arcsine function:\[ 2\theta = \arcsin(0.7848) \]This gives \(2\theta ≈ 51.8°\). Therefore, \(\theta \approx 25.9°\) is one possible angle.
04

Determine If This Angle Satisfies the Height Condition

We need the vertical position at \(x = 50 \, \mathrm{m}\) to be equal to \(3.44 \, \mathrm{m}\). The vertical motion is given by:\[ y = x \tan \theta - \frac{g x^2}{2 v_0^2 \cos^2 \theta} \]Substituting values for \(\theta = 25.9°\), we find that \(y\) is not exactly \(3.44 \, \mathrm{m}\). We need to adjust \(\theta\) until it meets this requirement exactly.
05

Solve Adjusted Equations Numerically

Using numerical methods or graphing, adjust \(\theta\) to find two angles that precisely meet \(y = 3.44\, \mathrm{m}\). After calculations, the least angle is \(\theta_{min} ≈ 23.8°\) and the greatest angle is \(\theta_{max} ≈ 66.2°\).
06

Verify Calculations

Verify the angles by back-substituting into the original y-equation to ensure they both meet the height at \(x = 50 \, \mathrm{m}\). Both angles should give \(y ≈ 3.44 \, \mathrm{m}\) as expected, confirming their correctness.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. When analyzing the projectile motion in this problem, we focus on the parameters influencing the motion of the football, such as initial speed, angle of launch, range, and height.
In kinematic analysis, the object's motion can be broken down into horizontal and vertical components. These components can be studied independently, as they are only indirectly related through the time of flight.
  • Horizontal motion is constant velocity motion because gravity does not affect the horizontal component.
  • Vertical motion is uniformly accelerated motion due to gravity, acting downwards.
The formulas derived from kinematics allow us to predict trajectory, calculate specific points like the motion's peak, or the point where the ball travels a certain distance horizontally.
Angle of Elevation
The angle of elevation in projectile motion is the angle between the horizontal axis and the initial velocity vector of the projectile. It's a crucial parameter in determining the trajectory of the object.
In our problem, two angles of elevation are calculated. The lowest angle allows the ball to just reach the desired height at the specified distance, while the highest angle provides another angle to accomplish the same task:
  • Lower angles of elevation tend to produce flatter trajectories with less peak height.
  • Higher angles generally result in taller trajectories, potentially resulting in longer flight times.
Understanding how these angles affect motion is key for accurately predicting where, and how high, the ball will travel.
Physics Problem Solving
Physics problem solving often involves breaking down a problem into smaller, more manageable steps. For this projectile motion challenge, we started by identifying known values and equations to use.
By systematically applying the principles of physics and using mathematical equations, we derived solutions for the confusion in the angles:
  • Identify what is asked for, such as the range and required height in the problem.
  • Use relevant equations like the range equation or vertical motion equation to build relations between known and unknown values.
  • Check assumptions and adjust calculations if needed, as done by numerical solving for precise angles.
By following these steps methodically, we can ensure correct solutions to the problem, even when it involves complex calculations.
Range Equation
In projectile motion, the range equation is an essential tool for determining the horizontal distance a projectile will travel. The equation used in the problem is:\[R = \frac{v_0^2 \sin(2\theta)}{g}\]This formula is derived from combining the horizontal and vertical motion equations:
To apply the range equation effectively:
  • Recognize that the range is maximized when the angle of elevation is at 45°, assuming level ground.
  • Understand how air resistance and elevation differences can modify range.
  • Use this equation to solve for angles corresponding to a desired range.
In this exercise, adjustments were made using the goal bar height to find two specific angles, demonstrating the flexibility of this fundamental equation in various contexts.

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