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How to find the initial velocity given only angle and distance traveled

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    An object is shot (from a cannon) at an angle of 33 degrees and landed 85 m away. Calculate the magnitude of the initial velocity (Hint:Look at the x direction and solve for Vox)


    2. Relevant equations
    Other questions I must answer. If you have time, help with these would be great, however I think if I can just find the initial velocity I should be fine.
    -Calculate the maximum height (***Actually, help with this one would really be nice!)
    -Calculate the time it took from launch to land (Easy enough)

    3. The attempt at a solution
    Well, I looked a a few different things around the internet and this forum, and I found one thing, maybe....
    So, I found the equation s=Vo^2/g*sin(2*theta). Using this equation, solving for Vo, I got 27.586 m/s. This seems like a reasonable enough answer, however I have absolutely no idea if it is correct.
    I may have approached it completely wrong, I really don't know....

    Thanks for the help! This is due tomorrow, and I really don't understand it.
     
  2. jcsd
  3. Sep 19, 2010 #2

    rock.freak667

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    Yes that formula is correct. The '85 m' is called the range, essentially how far it travels in the 'x' direction.

    Range = v02sin2θ/g

    To find the maximum height, answer this, at the highest point, what would be the vertical velocity? (rememeber, after this point, it starts to move downwards)
     
  4. Sep 19, 2010 #3
    Really, so 27.586 m/s would be correct, then, assuming I solved correctly for Vo? Great.

    Yes, so the object is at it's maximum height when V=0. V is decelerating at 9.8 m/s due to gravity. Now, I know that the angle, 33 degrees, is somehow involved here, no? But then what would my equation be =/ Umm... sorry, I am having trouble wrapping my head around an equation to use.
     
  5. Sep 19, 2010 #4

    rock.freak667

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    Right so vertically final velocity is zero. We have the initial velocity as 27.586 m/s at an angle of 33°, so what is the vertical component of this velocity?

    We have
    1) Final velocity
    2) Initial velocity
    3)Acceleration
    and we want to find 4) distance.

    Do you know a kinematic equation (SUVAT equations) that has all these quantities?
     
  6. Sep 19, 2010 #5
    V2=Vo2+2a(x-xo) ??

    If V is 0, Vois 27.586, a is g, X is our unknown, and Xo is 0.

    So, 0=27.5862+(2)(9.8)(X)
    So, 27.5862 / (2)(9.8)= X
    Correct?
    Then, 38.826 would be the answer. I think I got it. Thanks :D
     
  7. Sep 19, 2010 #6

    rock.freak667

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    Yes, that would be correct. But your equation should be (for projectile motion)

    v2=v02-2g(x-x0)

    That will take care of the negative sign when you rearrange and solve for 'x'.
     
  8. Sep 19, 2010 #7
    Ah, oops. Technicalities always seem to mess me up, haha. Thanks. This forum seems really cool. My physics teacher doesn't seem to know how to teach, so I'll probably be here a lot in the future, haha. Thanks again, you really saved me.
     
  9. Sep 19, 2010 #8
    Hmm... do you think I could have help with that last question that was supposed to be easy enough, Calculate the time it took from launch to land? Sorry, but I am a bit stuck.
    So, I would use the formula X=Xo+Vot+.5at^2, correct? But then, I get a bit confused...
    -.5t^2=Vot-X
    divide by t
    -.5t=Vo-X
    t=-2(Vo-X)
    but that is incorrect, I think... SO I dunno... Help, please... Thank you

    Oh, wait... Am I maybe just way over-analyzing this? Would it be as simple as Distance/Velocity? Therefore, V=30.197m/s (I made a mistake before. That is the correct V), and D=85m, so 85/30.197=2.815s. But that seems a bit short. The empirical data is 4s (we must find the percent error), that is quite a large percent error...
     
    Last edited: Sep 19, 2010
  10. Sep 20, 2010 #9

    rock.freak667

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    Right. We need to consider vertical motion to get this time.

    If it starts at x0=0, then when it lands, the vertical displacement will be x=0. So put x=0 and solve for t.
     
  11. Sep 20, 2010 #10
    Yeah, I got it after a bit. Thanks for all the help. Sure I'll be back soon with more obvious physics questions, haha.
     
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