Checking my work

  • Thread starter Chris-P
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  • #1
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Need Confirmation

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:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

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:
 
  • #3
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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
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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Hi Chris-P! :smile:

(just got up :zzz: …)

was that 802/2*9.81 ?

convert 80 ft to metres :wink:
 

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