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Finding Maximum Height and Initial Velocity when only given angle and distance

  1. Feb 1, 2011 #1
    1. The problem statement, all variables and given/known data

    A Circus performer is shot our of a cannon and shot over a net that is placed horizontally 6 meters from the cannon. When the cannon is aimed at an angle of 40 degrees above the horizontal, the performer is moving in the horizontal direction and just barely clears the net as he passes over it. What is the muzzle speed of the cannon and how high is the net?

    2. Relevant equations
    I have no idea, which is the problem

    3. The attempt at a solution

    assuming that R (sub) x is 6 I tried using the tan to solve for the hyp and then the maximum height but that was the wrong answer

    I can't seem ti figure out how to find the velocity becasue all of the equations i have need the time
  2. jcsd
  3. Feb 1, 2011 #2


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    welcome to pf!

    hi uncertain22! welcome to pf! :wink:

    in all these projectile problems, you need to do equations for the x and y directions separately (with accelerations 0 and -g, respectively) …

    call the time "t", choose one of the standard constant acceleration equations for each direction, and solve …

    what do you get? :smile:
  4. Feb 1, 2011 #3
    I tried doing it in the x and y directions seperatley but I don't know how to figure out the Velocity in the x and y directions without having an original velocity to go with.

    I tried doing the distance in the x and y directions in case that would help me somehow but it just kept coming out wrong

    Also, what does u stand for in the acceleration equations?
  5. Feb 1, 2011 #4


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    hi uncertain22! :wink:
    call the initial speed v …

    now show us your equations :smile:

    (u in s = ut + at2/2 is the initial speed in that direction)
  6. Feb 1, 2011 #5
    v^2 = (v)(cos 40) + 2 (-9.8) (6)

    i don't think that is right at all though
    and I don't know if 6 is even correct I feel like I need to put it into x and y components I just don't know how
  7. Feb 1, 2011 #6
    woops I think the (v times the cos of 40) is supposed to be squared too, I just forgot to type it in
  8. Feb 1, 2011 #7


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    (try using the X2 icon just above the Reply box :wink:)

    that's certainly a possible equation for the y-direction …

    what would it mean? what would v be in this case? why is it useful?

    (the 6 must be wrong, btw, it's a horizontal distance, and this is a vertical equation :redface:)

    try a different vertical equation if that one won't work

    (and then you'll need a horizontal equation also)

    (i'm going to bed now, so someone else will have to take over :zzz: …)
  9. Feb 1, 2011 #8
    how do I know which equations are vertical and which are horizontal?
  10. Feb 2, 2011 #9


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    hi uncertain22! :wink:

    (just got up :zzz: …)
    well, they're your equations … you can make whatever equations you want :smile:

    in the vertical direction, you need the standard constant acceleration equations, with a = -g

    in the horizontal direction, a is 0, so you can just use the standard constant velocity equation :wink:

    show us what you get :smile:
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