Check My Physics Homework: Ball Thrown Upward

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Homework Help Overview

The problem involves a ball thrown upward with an initial speed of 80 ft/sec, prompting questions about its maximum height and speed after 3 seconds. The subject area is kinematics, specifically dealing with projectile motion under the influence of gravity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the maximum height and speed at a specific time using kinematic equations. Some participants question the clarity of the problem statement regarding the maximum height and suggest alternative equations for verification.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's calculations and suggesting additional equations. There is recognition of differing answers for maximum height, indicating a productive exploration of the problem.

Contextual Notes

Participants note the potential confusion in the problem statement and the need for unit conversions, particularly from feet to meters. There is also mention of the original poster's limited exposure to the relevant equations.

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Homework Statement


A ball is thrown upward with an initial speed of 80 ft/sec. How high does it go? What is its speed at the end of 3 sec? How high is it at that time?


Homework Equations


v=v0+at
x=x0+v0+[tex]\Delta[/tex]t+.5a[tex]\Delta[/tex]t2

The Attempt at a Solution



Note: Only the first equation was taught to me, my teacher isn't the best...

Because it is in ft/sec I convert it and get 24.38 m/s rounded. So..
v=24.38-9.8t [-9.8 m/s because gravity right?]
So, in three seconds the velocity is -5.02 m/s.

So, I am trying to find the max height and that occurs at 0 m/s, so I plug in 0 for final velocity.

0=24.38-9.8t
t=2.49 seconds

Then, I use the second equation:
x=0+[24.38*2.49]+0.5[-9.8*2.492
x= 30.325 meters is the maximum height.

This is new territory for me, so I'd like to make sure I am correct. I have finals in a few days.
 
Last edited:
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Hi Chris-P! Welcome to PF! :smile:

That looks ok.

But there are three standard constant acceleration equations to learn.

To find the maximum height (was that part of the question? :confused:) you should use vf2 = vi2 + 2as :wink:
 
Thanks for the welcome :)

The question was a bit unclear as it says "how high does it go" so I assumed maximum height.

Also, thanks for the equation, but when I plugged everything in I got a different answer of 326.5 meters for maximum height :P
 
Hi Chris-P! :smile:

(just got up :zzz: …)

was that 802/2*9.81 ?

convert 80 ft to metres :wink:
 

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