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

Click For Summary
SUMMARY

The maximum height of a baseball thrown at a 75-degree angle with an initial velocity of 81 feet per second is calculated using the parametric equations for projectile motion. The vertical position is modeled by the equation y = -16T^2 + (81sin 75)T, where T represents time. The calculated maximum height is 95.62 feet, achieved at T = 2.4 seconds. This solution is validated through substitution from a similar example found in a math textbook.

PREREQUISITES
  • Understanding of projectile motion equations
  • Familiarity with parametric equations
  • Basic knowledge of trigonometric functions (sine and cosine)
  • Ability to perform calculations involving gravity in imperial units
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn about the effects of different launch angles on projectile height
  • Explore the concept of maximum height in projectile motion using calculus
  • Investigate the relationship between initial velocity and maximum height in various contexts
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in the mathematical modeling of sports trajectories.

dare2dream
Messages
14
Reaction score
0

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?
 
Physics news on Phys.org
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...
 

Similar threads

Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
12K
Replies
4
Views
5K
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K