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Projectile Motion problem

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    A ball thrown horizontally at 24 m/s travels a horizontal distance of 49 m before hitting the ground. From what height was the ball thrown?

    2. Relevant equations



    3. The attempt at a solution

    I'm not sure how to start this. Could someone please help?
     
  2. jcsd
  3. Sep 12, 2011 #2

    cepheid

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    You know how long it was in the air, because you have both the horizontal speed (which is constant) and the horizontal distance travelled. From this time spent in the air, you can deduce what height it must have fallen from (because in the vertical direction, the ball is just in free fall).
     
  4. Sep 12, 2011 #3
    Ok, I got the time. Now is there a specific equation I should use? I'm guessing it's just a variation of one of the Kinematic Equations?
     
  5. Sep 12, 2011 #4

    cepheid

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    Free fall means under the influence of gravity only. Hence acceleration is constant. This is exactly the condition in which the kinematics equations apply. To choose the correct one, consider your givens (in the vertical direction). You have been given the time, acceleration, and initial velocity, and you need to solve for the distance.
     
  6. Sep 12, 2011 #5
    Thank you so much! I've just got one more question: Since we are considering the vertical direction, acceleration due to gravity is still -9.8, not +9.8, correct?
     
  7. Sep 12, 2011 #6

    cepheid

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    It's up to you. You can pick either sign convention: "downward is negative," or, "downward is positive," so long as you stick to it consistently throughout the problem.
     
  8. Sep 12, 2011 #7
    Ok. I thought I did this correctly, however when I entered my answer of 28.6 it said that it was incorrect. I used d = vi*t + .5*a*t^2, where vi = 24, t = 2.04, and a = -9.8. Do you know where I went wrong in my calculations?
     
  9. Sep 13, 2011 #8

    cepheid

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    It doesn't seem like you're quite thinking things through here. Like with any projectile motion problem, we can consider the horizontal and vertical motions independently. I'll refer to the horizontal position coordinate as 'x' and the vertical one as 'y' to distinguish the two distances. This is really standard notation. For the x-direction, there is no acceleration, since the only force that acts is gravity, and it acts entirely vertically (in the y-direction). Hence, horizontal speed is constant, and the equation for distance vs. time is:

    x = vxt

    where vx = 24 m/s (given).

    You used this to solve for t, which was the total travel time. (If it had been in the air longer, it would have gone farther horizontally, and x would be correspondingly larger).

    In the y-direction, there IS acceleration, due to gravity. Hence, we can write ay = -g, where g = +9.81 m/s2, and I have chosen downward to be the negative y-direction. So, the formula for distance vs. time in this direction is:

    y = viyt - (1/2)gt2

    The initial vertical velocity (viy) is NOT 24 m/s (this is the horizontal velocity). However, you know what viy is from the situation. Hint: the thrower throws the ball entirely horizontally. This means that at the instant the ball is released, its only velocity is horizontal.
     
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