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Ball in free fall

  • Thread starter kasse
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  • #1
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A ball is shot out from a point A with initial velocity v. If it goes 55 meters in 4.4 seconds before it lands, what was the initial velocity and in what direction was the ball kicked?


I first find the initial velocity in x-direction: 12.5 m/s. Then I use energy conservation in z-direction to find an expression for the max height of the ball: v^2(initial in z-direction)/(2g)

This equals v(initial in z-direction)*t + (1/2)a*t^2.

But I don't get the correct answer. Where's my mistake?
 

Answers and Replies

  • #2
Hootenanny
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Your method thus far is correct. I can only assume that you have incorrectly solved for vz.

As an aside this isn't entirely a free-fall problem.
 
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  • #3
Redbelly98
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A ball is shot out from a point A with initial velocity v. If it goes 55 meters in 4.4 seconds before it lands, what was the initial velocity and in what direction was the ball kicked?
I'll Suggest another method to solve this. If the ball landed after 4.4 seconds, how long did it take to reach it's maximum height? Can you use that information to find the initial vz?
 
  • #4
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Yeah, that was easy, thanks!

I'm gonna have a small test soon. Is it enough to memorize these two equations for this kind of problems?

1) v = v0 + at
2) s = s0 + v0t + (1/2)at2

or do I need more equations?
 
  • #5
Redbelly98
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Yeah, that was easy, thanks!

I'm gonna have a small test soon. Is it enough to memorize these two equations for this kind of problems?

1) v = v0 + at
2) s = s0 + v0t + (1/2)at2

or do I need more equations?
There are 2 more:

3) [tex]
v^2 = v_0^2 + 2a(s-s_0)
[/tex]

4) [tex]
\frac{v+v_0}{2}= \frac{s-s_0}{t}
[/tex]

This last equation is just a statement about average velocity. It is omitted in some (many?) textbooks, but is just as useful as the others.
 

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