Kinematics of a kicked football

AI Thread Summary
The discussion focuses on solving the kinematics problem of a kicked football using the equation x = vcos(theta)t to determine the time to reach the goal post. Participants emphasize the necessity of including the vertical position equation to accurately assess the projectile's height at the goal post's elevation of 3.05 m. One user suggests using the trajectory equation and substituting specific values to find the correct answer, while also noting that significant figures may render the provided options incorrect. The conversation highlights the importance of both horizontal and vertical motion equations in solving projectile motion problems. Accurate calculations require careful consideration of all relevant equations and parameters.
Josh0768
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Homework Statement
Please help, I am thoroughly stumped trying to solve this kinematics question on my study guide. I’ve tried everything I know of but cannot get an answer that’s even an option.
Relevant Equations
R = v^2sin(2theta)/g
v^2 = vi^2 -2gd
x = vcos(theta)t
CE4D7DFC-357D-4EE6-8509-0FA17B8B104A.jpeg
 
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Josh0768 said:
x = vcos(theta)t
You could solve that equation for the time it takes to reach the horizontal position of the goal post.
It seems to me that you have forgotten to write down the equation for the vertical position of a ballistic object as a function of time.
 
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Josh0768 said:
I’ve tried everything I know
I assume you would like us to tell you where you are going wrong. Just a moment while I get out my crystal ball...
 
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Hello Josh.

How are you going to solve this problem without the equation for vertical position as a function of time? If you can figure out the elapsed time when the football is at the required height (3.05 m.) I think you can you then solve the rest of the problem.

AM
 
Josh0768 said:
Homework Statement: Please help, I am thoroughly stumped trying to solve this kinematics question on my study guide. I’ve tried everything I know of but cannot get an answer that’s even an option.
Homework Equations: R = v^2sin(2theta)/g
v^2 = vi^2 -2gd
x = vcos(theta)t

View attachment 249715
according to me the answer should be "G". Use equation of trajectory and put x=32.3 v=22.2 and theta=49.4 after plugging in the values you should get some "y" after that just subtract the height of the goal post.
 
If significant figures are to be taken into account, none of the answers are correct.

AM
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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