Need help on an object acceleration homework problem

  • Thread starter Physicshw
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  • #1
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Homework Statement



Someone throws an object upwards at 5 m/s from a window twelve feet high.

A) what is the velocity of the pen at 1 second after it was thrown?

Homework Equations



D = vi * t + (at^2)/2
Vf^2 = vi^2 + 2ad


The Attempt at a Solution


I keep getting 4.8, but the answer is 1.2m/s. What am I doing wrong???
 
Last edited:

Answers and Replies

  • #2
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Might want to edit your post and delete the unnecessary copies of the template :P

Anyway, when you throw something upwards at a given speed, it means its speed will decrease by g m/s every passing second.

If the window is 12 feet high - what is that in metres - 3.66 metres or so.
So the height relative to where you were throwing it is not 12 metres, but 12-3.66.
But there is a problem with the assignment - the ball at this small initial velocity cannot reach 8.34m height from the window if thrown up vertically. And it pretty much is showing in the numbers, your speed will decrease 9.8m/s every second, your initial speed is only 5 m/s, it means after 1 second it's already falling back down.

I calculated the minimum speed at which you will have to throw the ball and it is ~12,785m/s for it to even reach 8.34 height from the window.

Am I reading the assignment wrong? It is asking for the object's speed at 12 metres height from the ground with a given initial speed isn't it?
 
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  • #3
Ibix
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I, too, get 4.8m/s (which way?) based on the information you've given. I note that your first sentence says "object" and your second says "pen" - is that just a typo, or is it possible that you are getting information about different objects mixed up?
 
  • #4
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Oops, meant to say 12 meters.

And sorry for the template copies, I think it may be a glitch with the forum. I think it happened because I tried submitting the thread several times before I realized one of my tags was invalid.

Anyway, thanks for the reply.
 
  • #5
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Yeah, sorry about the typos, I'm in kind of a rush.
 
  • #6
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So, would -4.8m/s be correct? Maybe I wrote down the wrong answer...
 
  • #7
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I am getting -4.999 m/s considering the 12 metre height to be ground 0 so
5t - gt²/2 = 0 and v = v0 - gt
Probably rounding difference, but yeah, it cannot be 1.2m/s
 
  • #8
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Alright, thanks. that was driving me crazy for awhile.
 

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