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Projectile motion- solve for initial velocity

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown upward at an angle of 30degrees to the horizontal and lands on the top edge of a building that is 20m away. The top edge is 5.0m above the throwing point. How fast was the ball thrown?

    2. Relevant equations
    x=v0t+(1/2)at2
    vf2=V02+2ax
    vx=v0cosθ
    vy=v0sinθ

    3. The attempt at a solution
    I'm stumped on this question because I feel like I should solve for time first to get initial velocity but I can't figure out how to do that without actually having the initial velocity. I solved for t using t=20/v0cos30 and then plugged that into the vf2 equation but it comes out as a huge mess.

    I know the answer is 20m/s but I don't know how to get the answer.

    Thanks for any help!
     
  2. jcsd
  3. Feb 22, 2015 #2

    mfb

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    Staff: Mentor

    Your equation for x looks odd, should that be y? And what about the other coordinate?

    You have two conditions - the height and width - and two unknowns, time and initial velocity. Can you write down two equations that just depend on those (and the known angle)?
     
  4. Feb 22, 2015 #3

    TSny

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    Hello and welcome to PF!

    I don't see how you could substitute for t in the vf2 equation. That equation doesn't contain t. But you have the right idea. You need to substitute for t in the appropriate equation where you can use your information about the vertical height y.

    [Sorry. I see mfb posted while I was constructing my post.]
     
  5. Feb 22, 2015 #4
    You're definitely on the right track, but vf2 doesn't involve time, does it? Basically, you've solved for the time it takes for the ball to go 20 meters with initial velocity v0. Now, you also know that the ball has to travel 5 meters vertically in that same amount of time. Can you use that information to solve for v0 by plugging what you have for t into an equation for the vertical motion?
     
  6. Feb 22, 2015 #5
    I used t=20/v0cos30 and plugged that into t in the equation: vfy=v0y+ayt so it becomes vfy=v0y+ay(20/v0cos30) Then I found vfy to be (v0y2-(2*-9.8*5))1/2 and plugged that whole thing into the vf=vo+at formula. So: (v0y2-(2*-9.8*5))1/2=v0y+(-9.8)(20/v0cos30) That way the only variable is v0y but I don't know where to go from there
     
  7. Feb 22, 2015 #6
    Is there a simpler way to solve this that I'm missing? In my practice book it is only a level 2 problem (out of 3) but it seems very complicated
     
  8. Feb 22, 2015 #7

    PeroK

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    Trying to find the final vertical velocity is not necessary. Instead, think about the y displacement.
     
  9. Feb 22, 2015 #8
    I'd say step back and think about what you're trying to solve for. You want v0. So you have two unknowns really: time and the initial velocity. When you have two unknowns you're going to need two equations that involve those two unknowns to solve for them both. You're making things more complicated by involving a third equation, I think.

    You can use the initial and final y velocity equations to solve I believe, but remember you need everything to be in terms of v0 in order to solve for v0. OR, you need everything in terms of v0y. You have equations relating the two, so you can do either by subsituting.

    PeroK's hint is correct though, this problem is much easier if you simply use two equations of motion for displacement, one in the x direction and one in the y direction.

    This might make things a little clearer

    y = v0yt+1/2at2
    x = v0xt

    Use:
    v0y = v0 sinθ
    and
    v0x = vo cosθ

    To put everything in your displacement equations in terms of v0 and the rest is algebra (which might be a little tricky, but if you write down all your steps shouldn't be too bad).
     
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