Conservation of Energy Problem.

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  • #1
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Here's the problem:

A .40 kg ball is thrown with a speed of 12m/s at an angle of 33 degrees. a) what is its speed at its highest point, and b) how high does it go? (Use conservat ionfo energy and ignore air resistance)

I think I'm supposed to use the formula .5mvi^2 + mgy1 = .5mvf^2+mgy2 .

How do I find the speed at the highest point? m=.4, and vi=12 right? Please help!
 

Answers and Replies

  • #2
arildno
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A couple of hints:
What is the vertical velocity at its highest point?
Is the horizontal velocity at all affected during the flight?
 
  • #3
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At the highest point, vertical velocity is 0, and no, the horizontal velocity isn't affected... so then my final velocity would be zero, and y1 will be zero. This leaves me with 28.8-20 = (.40)(9.8)y2
y2 = 2.21 meters. Can anyone verify my answer?
 
Last edited:
  • #4
Pyrrhus
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Looks ok to me.
 

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