What Is the Maximum Height of a Baseball Thrown at 75 Degrees?

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Homework Help Overview

The discussion revolves around estimating the maximum height reached by a baseball thrown at an angle of 75 degrees with an initial velocity of 81 feet per second. The problem is situated within the context of projectile motion in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of parametric equations to model the baseball's trajectory, with one participant expressing uncertainty about their calculations and seeking verification. Questions arise regarding the application of equations and the significance of the vertical component of velocity at the peak height.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the equations used. Some guidance has been offered regarding the physics principles involved, but there is no clear consensus on the correctness of the initial calculations or the approach taken.

Contextual Notes

Participants mention reliance on examples from a textbook, which may not provide comprehensive explanations of the underlying physics. There is also a noted lack of support from the instructor, contributing to the participants' uncertainty.

dare2dream
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Homework Statement


Estimate the maximum height reached by a baseball during its flight if it is thrown with a velocity of 81 feet per second at an angle of 75 degrees relative to level ground.


The Attempt at a Solution


I found an answer, but I don't know if I did it right so I guess I'm just asking for someone to check it.

I found x = (81cos 75)T and y = -16T^2 + (81sin 75)T
So I put that in my calculator in parametric mode and graphed it with the right window. I got that the highest it would reach is 95.62ft when T=2.4

Is that right?
 
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The answer you got doesn't look right. Could you show how you got that result?
 
<.< I used my calculator with that information, guessed at different T values and found that when T was 2.4, it was the highest?

I don't know what else to show you...
 
Didn't you use any equations? What does the y-component of the velocity equal when the ball is as its highest point? What does that tell you about the time?
 
What are you talking about? I gave you the equations I used...
 
dare2dream said:
What are you talking about? I gave you the equations I used...

dare2dream said:
y = -16T^2 + (81sin 75)T

Where did you get this from?
 
That's gravity...or something.

There was a similar example my math book and I just substituted the different angle and velocity.
 
I still don't know if that's right or what to do/how to fix it if it's wrong...
 
  • #10
dare2dream said:
I still don't know if that's right or what to do/how to fix it if it's wrong...

Have you read the page that radou provided in the link above. Again, how did you come up with the equation for the vertical distance?

Also, I would strongly recommend against using a calculator to perform these calculations for you. You should be able to do it by hand; and if not, now's a good time to learn how to!
 
  • #11
Yes, I read the link. I found the equations by an example in my math book.

This is the example:
Kevin hits a baseball at 3ft above the ground with an initial speed of 150ft/sec at an angle of 18 degrees with the horizontal. Will the ball clear a 20-ft wall that is 400 ft away?

The path of the ball is modeled by the parametric equations x = (150 cos 18)t, y = -16t^2+(150 sin 18)t + 3

---
I used that example and substituted the numbers. That's the only way I could figure out the problem.
 
  • #12
The answer you found seems right. I don't know how detailed a solution you are looking for, if you don't know the physics behind the motion. Does your book not explain it?
 
  • #13
dare2dream said:
Yes, I read the link. I found the equations by an example in my math book.

This is the example:
Kevin hits a baseball at 3ft above the ground with an initial speed of 150ft/sec at an angle of 18 degrees with the horizontal. Will the ball clear a 20-ft wall that is 400 ft away?

The path of the ball is modeled by the parametric equations x = (150 cos 18)t, y = -16t^2+(150 sin 18)t + 3

---
I used that example and substituted the numbers. That's the only way I could figure out the problem.

Oh I see, your book uses imperial units (for some strange reason), so that would explain why g=16!

Still, I stand by my comment that you should not rely on a calculator for this problem. Surely you must have done some theory before you come across the example in the textbook.
 
  • #14
That's all the my book explains...so it's not all that helpful. Our teacher isn't that helpful either. =S

If you can do it another way and get the same answer, that's good enough that I'll leave my work how it is...
 

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