What is the maximum height of the stadium at its back wall?

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    2d Kinematic
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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a baseball hit at a certain velocity and angle. The objective is to determine the maximum height of a stadium wall that the baseball would just clear, given its initial conditions and distance from the point of impact.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations involved in determining the time of flight and the vertical height of the baseball at a specified distance. There are attempts to verify the accuracy of the calculations and the values used in the equations.

Discussion Status

Some participants have confirmed that the steps taken appear to be correct, while others have noted minor discrepancies in the numerical values. There is an ongoing exploration of the calculations, with one participant considering seeking clarification from a professor regarding potential grading issues.

Contextual Notes

Participants are working under the constraints of an online grading system that may be sensitive to numerical precision. The problem setup includes specific initial conditions and assumptions about air resistance.

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Homework Statement


Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of θ = 35 degrees to the horizontal. The ball is struck 3 feet above home plate. You watch as the ball goes over the outfield wall 420 feet away and lands in the bleachers. After you congratulate Ted on his hit he tells you, 'You think that was something, if there was no air resistance I could have hit that ball clear out of the stadium!'

Assuming Ted is correct, what is the maximum height of the stadium at its back wall x = 565 feet from home plate, such that the ball would just pass over it?

You may need:
9.8 m/s2 = 32.2 ft/s2
1 mile = 5280 ft

Homework Equations


Xf=Xi+Vit+0.5a(t^2)

The Attempt at a Solution


Horizontal:
a=0
Vi=176cos35
xi=0
Xf=565

Vertical:
a=-32.2ft/s^2
Vi=176sin35
Xi=3 ft

i simply plugged the horizontal values in and solved for t which was 3.918

i then used t and the vertical components and simply plugged it in and gave me Xf being 151.34ft ft which is wrong... This is a online grading so they might be picky about the accuracu but i just want to be sure I'm doing the steps right.
 
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The steps seem OK. The values are off in the last digit: I get 3.919 and 151.35.
 
Why are you bumping this?
 
if voko's post didn't help, then you should explain why :)
 
woops wrong post i bumped LOL. But yea... still giving me me as the wrong answer for some reason... i think i'll just ask my proff if there's something wrong with the grading system
 

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