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Homework Help: Basic projectile motion with trig

  1. Nov 10, 2011 #1
    A field goal kicker kicks the football so that it leaves the ground with velocity V that is 50 degrees above horizontal. The ball travels for 3.24s and hits the crossbar 10.00 ft above the ground. Neglect air resistance.
    Find the velocity after the kicker kicks the ball.
    Find the horizontal distance the ball was from when kicked.

    Ok, I wrote all the basic equations of motion. All my attempts to combine the equations through substitution netted me too many variables and I couldn't solve anything. I really don't understand where I'm going astray. This problem wouldn't be very difficult if the ball wasn't hitting the crossbar above the ground. Also, finding the distance would by child's play if I could just wrangle the initial velocity out of this thing.

    Edit: This isn't really a homework problem, it's on a study guide and I'm almost certain I'll see something like it on my exam tomorrow. I'm not looking so much for a solution but a general strategy that lets me attack projectile problems like this when the projectile doesn't land at y=0.
    Last edited: Nov 10, 2011
  2. jcsd
  3. Nov 10, 2011 #2
    Having too many variables is likely the point of the exercise. Have you tried picking some approximate values for the initial velocity to find the range wherein it lies? A spreadsheet makes this a relatively simple task.
  4. Nov 10, 2011 #3
    I'm going to be expected to do this with just a scientific calculator. I know there is a simple way to figure this out that is just flying right over my head.
  5. Nov 10, 2011 #4


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    Firstly deal with the vertical component - we can convert to the actual velocity later.
    First list the variables:
    Vo = ??? - I use the question marks as this is what we are after.
    x = 10
    t = 3.24
    a = -9.8

    we need the equation that has no Vf in it.

    x = Vo.t + 0.5.a.t2


    10 = Vo x 3.24 - 4.9 x 3.242

    re-arrange and solve for Vo
    Remember this is just the vertical component, you have to undo some trig on the 50o angle to find the actual velocity.

    I reckon you should be able to do it from there.
  6. Nov 10, 2011 #5
    I don't agree. All that equation says is that in 3.24 seconds the ball travels up 10 ft from zero. What's happening is the ball is traveling up to a height y2 then down to 10ft. Were also in ft here so it's easier to use ft/s2 for the acceleration.

    Anyway my exam is in 5 minutes so, hopefully I can figure it out if it shows up.

    edit:Test done, went well, this didn't show up. Still interested in an answer.
    Last edited: Nov 10, 2011
  7. Nov 10, 2011 #6


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    The mistake I made was not converting the 10 ft to metres - we dropped feet etc in 1974.

    The beauty of the motion formulas is that they include displacement - so all that is important is that the object [the ball] finishes its journey 10 ft [3.xxxxx m ] above where it started.

    So I should have adjusted the height to metres, or perhaps used the ft equivalent of g - I recall it is about 32???

    Once you have found the initial velocity, you could use that same formila, but use the Vo value and use just t to find when it was at height 10 ft and you will get not only 3.24 seconds, buta time much less that 1 second. That little time is on its way up.
  8. Nov 10, 2011 #7
    Oh wow, I didn't think the motion equations worked like that. Though, I guess being a second degree polynomial should have been a huge tip off.

    Problem solved.
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