Solving for Time and Max Height of a Kicked Football

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Homework Help Overview

The problem involves a football kicked towards a field goal, requiring the calculation of the time it takes to pass through the uprights, its maximum height, and its launch angle. The context is rooted in projectile motion, with specific parameters provided such as distance, initial velocity, and height at a certain point.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss breaking down the initial velocity into its horizontal and vertical components and explore the relationships between distance, time, and velocity. Questions arise regarding the complexity of the resulting equations and the best approach to simplify calculations.

Discussion Status

Some participants have suggested focusing on using variable representations instead of numerical values to simplify the equations. There is acknowledgment of the challenges faced in solving the equations, with expressions of frustration over complex calculations. The discussion remains open, with no explicit consensus reached on the solutions.

Contextual Notes

Participants note the difficulty in managing trigonometric equations and the potential for multiple solutions to the questions posed. There is a mention of prior class discussions on similar problems, indicating a shared learning experience.

kk727
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Question: 1. A football is kicked towards a field goal 28m away. The initial velocity of the ball is 37 m/s. The ball crosses safely through the uprights, while it is 3m above the ground.
a. Find time when it passes through the uprights.
b. Find its max height.
c. Find its launch angle.

Homework Equations


v = vo + at
x = xo + vot + 1/2at2

The Attempt at a Solution


So, we started this problem in class today, and didn't have enough time to solve it completely.
We first split the velocity into its X and Y components:
VoY= 37sinθ
VoX=37cosθ

Using V=x/t, we figured out that 37cosθ=28/t, and t=28/37cosθ.

Then using this equation, we plugged in everything that we had.
3 = 0 + 37sinθt - 4.9t2
3 = 0 + 37sinθ(28/37cosθ) - 4.9(28/37cosθ)2

We attempted to solve for t by plugging in 28/37cosθ for it, but ended up getting a bunch of trig, and I personally couldn't solve it. Can anyone help out? Is everything at least right up to here? Thanks in advance!
 
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Forget about plugging in the values of V_{oy}[/tex] and V_{ox}[/tex] for now ... just use the variables in your equations.<br /> <br /> For example, with V = x/t, try using V_{ox} = \frac{28}{t}[/tex]<br /> and go from there. Your calculations will be <b>much</b> easier!
 
Thank you so much! That's something that I've been struggling with for a while now, because we've been doing a lot of problems like that in class, and it frustrates me when I end up with this giant, ugly equation! You're right, it's muuuuuch easier!
 
kk727 said:
Thank you so much! That's something that I've been struggling with for a while now, because we've been doing a lot of problems like that in class, and it frustrates me when I end up with this giant, ugly equation! You're right, it's muuuuuch easier!

No problem!

You should have come up with 2 answers for each question. What did you get?
 
Oh, sorry, I forgot to check this again! I don't remember, because I don't have my notebook with me, but we went over it in class and I had it right! Everyone else had big long trig equations that they were trying to solve and I felt so smart! Hahaha, thanks again!
 

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