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Basic Kinematics 2D problem

  1. Oct 4, 2013 #1
    1. Problem
    The kicker stands 50 m from the goalpost that is 3 m high. She kicks with an initial velocity of 24.5 m/s. She just makes the goal height-wise, but did the ball travel far enough? She kicked at an angle of 35.28 degrees.

    2. Relevant equations
    S = (1/2)at^2 + V0t + S0
    S =V0xt (when V0x is constant)

    3. What I Did
    24.5cos(35.28) = V0x = 20
    24.5sin(35.28) = V0y = 14.15
    -5t^2 + 14.15t + 0 = 3
    t = 2.6
    20*2.6 = 52 m
    2 meters to spare

    The answer is 1.12 m to spare. Any help would be appreciated! Thanks.
    Last edited: Oct 4, 2013
  2. jcsd
  3. Oct 4, 2013 #2


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    Staff: Mentor

    What a strange question. Surely if the ball "just made the goal height-wise" the ball is still in the air when it passes over the goal post, the goal has been made, and by definition the ball has traveled far enough.

    From the given answer it looks like they meant to ask you by what margin the ball cleared the goal post, and g = 10m/s2 is assumed for the acceleration due to gravity.

    Have you stated the question exactly as given to you?
  4. Oct 4, 2013 #3
    a) gravity is assumed to be -10 m/s
    b) Kyle will attempt a goal from 50 m in front of the goal post. Kyle boots the ball with an initial velocity of 24.5 m/s at an angle of 35.28. It is right on target but is it long enough? How many meters to spare?
    Last edited: Oct 4, 2013
  5. Oct 4, 2013 #4


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    Staff: Mentor

    So it would appear that my interpretation was correct. Find the height of the ball when it passes the goal post. Compare with the height of the goal post.
  6. Oct 4, 2013 #5
    Hmmm the answer of 1.12 looks suspicious.
  7. Oct 4, 2013 #6
    I think it is asking me, length-wise, how much I have to spare, not height-wise; it asks me "is it long enough"?

    What makes you say the answer is suspicious?
    Last edited: Oct 4, 2013
  8. Oct 4, 2013 #7
    It is a little hard to understand what is being asked.
  9. Oct 4, 2013 #8


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    Staff: Mentor

    The question is not well posed.

    Length-wise, the velocity in the x-direction is (theoretically) constant ignoring air resistance and other losses. There's no limit to what lengthwise distance the shot can reach, even if it bounces while doing so. So it seems strange that they'd ask for a lengthwise distance limit unless there were some other criteria associated with it, such as the ball must not touch the ground before reaching the goal line. But nothing like that was specified. We don't even know what game is being played and what constitutes scoring a goal! What was given was a goalpost height and distance.

    If we consider typical questions of a similar nature, usually they want to know if the shot will be able to clear a certain height at a given horizontal distance. For the shot described in this problem it will clear the given goalpost height at the horizontal distance given for the goal line, and will do so by an amount very close to the given answer. In layman's terms that shot is "long enough" for a shot that must clear the goal height without bouncing first.

    You could calculate the range of the shot (distance from launch to hitting the ground for the first time), and that will definitely be further than the goal line if it's still more than 3m high when it reaches the goal line and lands at the same angle as it was launched (projectile trajectories are symmetric when air resistance is ignored).
  10. Oct 5, 2013 #9
    IIRC, the answer yields around 58 m; 8 m to spare. Intuitively, you can see that the range of X is greater than X when Y = 3 so it's not getting closer to 1.12 m.
    0t^2 + 20t + 0 = 50
    -5(2.5)^2 + (14.15)(2.5) + 0 = S
    S = 4.125 m; excess of 1.125 m
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