Finding initial velocity given angle and horizontal displacement

  • Thread starter Suprin
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  • #1
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Homework Statement


distance = 0.8meters
angle = 55°
gravity = [itex]9.81m/s^2[/itex]
[itex]\V_0[/itex] = ?, assuming that it leaves and falls back on the same horizontal plane.



The attempt at a solution

So we know for a fact that the initial velocity is supposed to be [itex]2 m/s[/itex], according to the book at least.

I tried using the formula [itex] R = \frac{V_0^2 sin2\theta}{g} [/itex] but that gives me a completely different number.

I've discarded the use of component formulas since I'd require initial velocity for those. I'd like to say that I believe that the initial velocity on X is different to the initial velocity on Y since it's not 45°, but I get the feeling that doesn't matter. I'm stuck and frustrated on this relatively simple problem and can't figure it out for the life of me.

I'm also trying to get the hang of this Latex thing :p
 
Last edited:

Answers and Replies

  • #2
6,054
390
The formula you have for range is wrong, which is obvious if you check the units. Use the correct formula and you should be fine.
 
  • #3
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If you meant the [itex]V_0^2[/itex], I fixed it. That was a typo. Still not getting the right answer.
 
  • #4
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Use the equations for the x and y positions for projectile motion and use the final positions for x and y.
 
  • #5
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If you meant the [itex]V_0^2[/itex], I fixed it. That was a typo. Still not getting the right answer.
Show exactly what you do.
 
  • #6
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Show exactly what you do.
I have no idea what I did wrong before, but as soon as I was about to finish copying the last step to reply to you, I got the answer.

It was one of those "watch which buttons you press on the calculator" things. Sorry :/

Rounding at the very end, I do get 2.9m/s.
 
  • #7
6,054
390
No need to be sorry, you solved the problem, and I am glad I was able to help, even if only by having you do that again :)
 

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