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Math Help - Motion of ball

  1. #1
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    Motion of ball

    I have a math homework question that reads "On Earth a ball is thrown upward from an initial height of 5 meters with an initial velocity of 100 m/s. How long will it take the ball to return to the ground? "

    I am so confused over this whole motion thing. I have tried using the equations of motion but I fail every time to produce a correct result .. thanks for any help.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Mathfailure View Post
    I have a math homework question that reads "On Earth a ball is thrown upward from an initial height of 5 meters with an initial velocity of 100 m/s. How long will it take the ball to return to the ground? "

    I am so confused over this whole motion thing. I have tried using the equations of motion but I fail every time to produce a correct result .. thanks for any help.
    We are given the following initial conditions:
    y0 = 5 m
    v0 = 100 m/s
    y1 = 0 ... since we want it to hit the ground, we need its 'final' height to equal 0.

    We'll use the distance formula:
    y = y0 + v0*t + 1/2*g*t^2 ... where y0 is the initial height of the ball, v0 is the initial velocity, g is the acelleration of gravity = -9.8 m/s^2:

    0 = 5 + 100t - 4.9t^2
    4.9t^2 - 100t - 5 = 0

    Now, use the quadratic formula to solve for t:
    t = (100 + sqrt(10098))/9.8 = 20.46 s

    NOTE: I did the positive version of the quadratic formula since the 'minus' version would give a negative number for t (which we don't need).
    Last edited by ecMathGeek; March 26th 2007 at 06:34 AM. Reason: I had the wrong units written for velocity
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  3. #3
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    That was amazing.. thanks...
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  4. #4
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    Another way to do it is to figure out the time it takes to reach the peak. On Earth, gravitational acceleration is 9.8 m/s^2. Each second the acceleration basically reduces the speed by 9.80 m/s.

    original velocity/9.80 = time to peak
    100/9.80=10.2 seconds to peak. A projectile fired off will produce a nice, symmetrical parabola (or, in this case, line straight up and down) where time from original height to peak = time from peak to original height and velocities at matching heights are equal. The time in the air to get back down is therefore 2*10.2 = 20.4s This leaves the projectile at 5 m in the air, traveling at 100 m/s.

    d=vt
    t=d/v
    t=5/100
    t=.02 s

    Total time, therefore is 20.4 + .02 = 20.4s. The differences between our answers can be attributed to rounding and significant figures (I chose to take into account sig figs, which your teacher may or may not have you do).

    -Pulsar
    Last edited by Pulsar06; April 2nd 2007 at 05:58 PM. Reason: lol, 20.4 + .02 =/= 20.6
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    We are given the following initial conditions:
    y0 = 5 m
    v0 = 100 m/s
    y1 = 0 ... since we want it to hit the ground, we need its 'final' height to equal 0.

    We'll use the distance formula:
    y = y0 + v0*t + 1/2*g*t^2 ... where y0 is the initial height of the ball, v0 is the initial velocity, g is the acelleration of gravity = -9.8 m/s^2:

    0 = 5 + 100t - 4.9t^2
    4.9t^2 - 100t - 5 = 0

    Now, use the quadratic formula to solve for t:
    t = (100 + sqrt(10098))/9.8 = 20.46 s

    NOTE: I did the positive version of the quadratic formula since the 'minus' version would give a negative number for t (which we don't need).
    This is a correct solution, but I wish to point out that when Pulsar06 put the numbers into his y = y0 + ... equation he assumed two things:
    1) The origin of his coordinate system is on the ground.
    2) The +y direction is upward.

    As we have complete freedom to pick a different origin and different positive direction I always recommend these two points be noted when doing any problem.

    -Dan
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