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

AI Thread Summary
The discussion centers on calculating the maximum height of a baseball thrown at a 75-degree angle with an initial velocity of 81 feet per second. The user derived the equations for horizontal and vertical motion but is uncertain about the accuracy of their results, which indicate a maximum height of 95.62 feet at T=2.4 seconds. Other participants emphasize the importance of understanding the physics behind projectile motion rather than relying solely on calculators. They suggest reviewing the equations used and the underlying principles of motion, particularly gravity's effect. The conversation highlights a need for clarity in applying physics concepts to solve the problem effectively.
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

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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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