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Projectile Motion: Finding the Maximum Height of a Football

  1. Nov 4, 2015 #1
    A football is kicked from a tee to start the football game. If the ball is kicked at a 38 degree angle with an initial velocity of 27.1 m/s, Find the maximum height of the football as it flies through the air.

    I was using these equations:
    Voy = Vo(sine of theta)
    Vy = Voy + at
    y = 1/2(a)(t^2)

    My attempt:
    Voy = 27.1(sin38)
    Voy = 27.1(.62)
    Voy = 16.68

    -16.68 = 16.68 + -9.8t
    -33.36 = -9.8t
    t = 3.4

    y = 1/2(-9.8)(3.4^2)
    y = -56.644

    Lon Capa only wants a less than three digit answer, and it keeps saying that it's wrong. I've tried -56 m, 56 m, -57 m, and 57 m. Please help.
     
  2. jcsd
  3. Nov 4, 2015 #2

    SteamKing

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    Everything seems OK to this point.
    What's the deal with the velocities here?
    What is the vertical velocity of the football when it reaches its max. height above the ground?

    Hint: It's not double its initial upward velocity. And why is the vertical velocity negative?
    Be careful with signs. Generally, a positive vertical distance means the ball is traveling above the ground.

    Who is Lon Capa? Is he someone who hangs around your school?
     
  4. Nov 4, 2015 #3
    I used the Voy that I got from the first equation, and then set Vy = -(Voy). But maybe that was wrong because I'm not looking for the final velocity of the motion, I'm looking for the velocity of the football at the max point? (is that what you were trying to say?) And the velocity at the max point is 0 m/s, correct?

    Lon Capa is a website that my AP Physics teacher uses to give us homework through. You answer the question by typing the answer into a box, and you get about 8 tries. When the box goes green, you're correct, but when it goes red, you're incorrect.
     
  5. Nov 4, 2015 #4

    SteamKing

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    Yes, the vertical velocity of the ball is zero when it reaches max. height.

    Adjust your velocity calculations to obtain the correct time it takes the ball to go from the kicking tee to this height.
     
  6. Nov 4, 2015 #5
    I did go back and set Vy = 0 m/s instead of -16.68 m/s in the second equation, and doing that, I got y to eventually equal 14. I entered the answer into Lon Capa and got it right!

    There is another part to the question, if you wouldn't mind helping with it.

    If the hang time of the ball (total time in the air) is 6.5 seconds, find how far the football goes.

    I tried to use x = Vox(t)

    Vox = 21.36 (which I found using Vox = Vo(cosine of theta), Vox = 27.1(sin38), Vox = 21.36)
    x = 21.36(6.5)
    x = 138

    When I entered this into Lon Capa, it said it was wrong because the number was too high.
     
  7. Nov 4, 2015 #6

    SteamKing

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    You've got one typo in your formula for Vox:

    Vox = 27.1 * cos 38 = 21.36 m/s

    Your arithmetic is correct in finding the distance the ball travels at this horizontal velocity if the total hang time is 6.5 sec. A more reasonable number is about 4.5 sec.

    However, since the ball only took about 1.7 sec to reach max. height, it's not clear why the total hang time is almost four times this duration.
    You should confirm the total hang time for the ball.

    BTW, in actual football play, a hang time of 6.5 sec is almost unheard of, unless Superman is your kicker.

    A kicker who is able to kick a football almost the length of a football field and a half would find most stadiums very confining.
     
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