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Motion in two dimensions question. At wit's end.

  1. Mar 26, 2008 #1
    [SOLVED] Motion in two dimensions question. At wit's end.

    Hi, I'll try to keep this short. I have a Physics exam tomorrow and was doing some textbook questions and can't figure out this one problem. It's killing me! I have the answer (supplied by the back of the book), but just can't seem to get there.

    Some help would be much appreciated since I've grown very frustrated with it.


    1. The problem statement, all variables and given/known data

    The initial speed of a projectile is 80 m/s. If the projectile is to strike a target that is a horizontal distance of 450 m away, what is the minimum time of flight. (Answer 6.1s)

    2. Relevant equations

    Standard laws of motion ignoring air resistance and the like

    X(final) = X(initial) + V(initial) (t) + 0.5at^2


    3. The attempt at a solution

    I've tried everything I can think of. What's shutting me down every time is that there is no angle supplied so I'm always stuck with two variables I can't eliminate.

    Obviously the answer must lie in finding the smallest angle possible for a projectile at 80 m/s to reach 450 meters, I just can't figure out how. I came up with this equation from the formula above, but I have no clue how to use it as my brain is just shutting down at this point (maybe it's useless i don't know, argh!)

    450 = 0 +80(cos(alpha))t = 0
    t= 5.625/cos(alpha)



    If someone could help me get this problem you would earn my undying admiration, and I'll mail you a cookie!
     
    Last edited: Mar 26, 2008
  2. jcsd
  3. Mar 26, 2008 #2

    berkeman

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    Staff: Mentor

    Take a step back. The target is 450m away, so the time it takes to get there is what you calculated in terms of alpha. The other constraint is that the projectile has to go up for half of the flight, and back down for the other half of the flight, in order to hit the target. That up and down motion has to take the same total time t.

    You'll probably get a 2nd order equation, since there are probably two answers (low angle shot < 45 degrees, and high angle shot > 45 degrees)
     
  4. Mar 26, 2008 #3

    Doc Al

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    Staff: Mentor

    You already have an equation for the horizontal position as a function of time. Now write an equation for the vertical position as a function of time. Combine those equations to eliminate the angle and solve for the time.
     
  5. Mar 26, 2008 #4
    Thanks for the help! I'm going to try it again and I'll post back, hopefully to add a solved tag , heh.
     
  6. Mar 26, 2008 #5
    Finally managed it. Thanks for the help guys. All I had to do was use the sin/cos identities to solve for alpha, I feel retarded, thanks again!
     
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