Finding Optimal Elevation Angles for a Field Goal: A Kinematics Trig Problem

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Homework Help Overview

The discussion revolves around a kinematics problem involving the optimal elevation angles for a football kicker to successfully score a field goal from a specific distance and height. The original poster presents an equation derived from the problem setup but expresses uncertainty about its solvability and the approach taken.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various trigonometric identities and substitutions to manipulate the original equation. There is a focus on transforming the equation into a quadratic form, with questions about the necessary steps to achieve this.

Discussion Status

Participants are actively engaging with the problem, suggesting different identities and transformations. There is no explicit consensus on the best approach yet, but several productive lines of reasoning are being explored.

Contextual Notes

The original poster questions the solvability of their equation without a graphing calculator, indicating potential constraints in their approach. There is also a discussion about the need to express terms consistently for solving the quadratic equation.

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Homework Statement


A football kicker can give the ball an initial speed of 25 m/s. What are the least and greatest elevation angles at which he can kick the ball to score a field goal from a point 50 m in front of goalposts whose horizontal bar is 3.44 m above the ground?

The Attempt at a Solution


I've worked it down to 3.44(cos a)^2 = 50(sin a)(cos a) - 19.6 , but I can't figure out how to solve this equation. None of the trig identities seem to help. Is this equation solvable (without a graphing calculator) or am I just approaching the problem the wrong way?
 
Last edited:
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try
2sin(a)cos(a) = sin(2a)
and
cos(2a) = 2cos(a)^2 - 1
 
Made the subs, now I have 1.72 cos(2a) - 25 sin(2a) = -21.32 . What now?
 
Last edited:
Assuming you did everything correctly, you can use the identity cos(2a)=1-2sin^2(a) to get a quadratic equation.
 
Wouldn't I have to either get everything in terms of (sin a) or everything in terms of (sin 2a) to solve as a quadratic?
 
You can subsitute cos(2a) = (1-sin^2(2a))^0.5 and square the equation to get a quadratic equation. Solve for sin(2a).
 

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