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Please, need help for a problem.

  1. Sep 22, 2006 #1
    Please help, 2D motion problem, projectile movement.

    Two dimensional motion with constant acceleration problem,
    please help this is the last one i have to solve and i can't come up with anything.

    Basketball player covers 2.80m horizontally in a jump.
    His center of mass moves through the space as following:
    his center of mass is at elevation 1.02m when he leaves the floor,
    it reaches a maximum height of 1.85 above the floor, and its at 0.900m when he touches down again.

    determine his flight "hang" time, angle of takeoff, horizontal and vertical velocity components at takeoff.

    Please help, i don't know where too start, seems like too much is unknown...


    I started by making a drawing, and i thought that since max height and range are given ill go from there, but the motion is not symmetrical

    So i wrote out the X and Y motion equations:

    Xf = Vix * t = Vi cosA * t , where A is the unknown angle V unknown speed, t unknows time
    Xy = Vyx *t = Vi sinA * t + 1/2gt^2

    But i cant solve systems because there are 3 unknowns and 2 equations

    So i went back to max height and range again, i tried to figure out what the range was for symmetrical movement, but i couldn't since i dont know the angle, and even if i did i wouldn't know what to do with the rest of the motion trajectory, so thats it, im stuck here, i don't even know what to do because it seems that to solve it i need either angle or initial speed..

    Help please...
    Last edited: Sep 22, 2006
  2. jcsd
  3. Sep 23, 2006 #2
    well first we'll start with what you do know...so you know that his total distance in the x direction is 1.8m lets call this x while his initial starting point in the x direction is 0 this will be x_0 so this is all the information for the x direction we are given so lets leave that and see if we can do anything in the y direction.

    we know tht y=1.85m we will ignore his height at the landing for this step... y_0 is 1.02m
    his initial velocity is 0 in the y direction so this will be v_0=0
    and his final velocity is also 0 at 1.85 meters y , the acceleration can be taken as -9.8m/s^2
    so now we have a good place to start by using the equation y=y_0+v_0t+.5at^22 this will give you the time in order for him to get this high... now you use the same equation and method to find the time it takes for him to reach the height of .9m...add these two times and you have the total time!!
    that should at least give you a start on the problem
  4. Sep 23, 2006 #3
    could u please elaborate a little? i dont quite get it,

    r u saying that vertical component of velocity at 1.85 max height point is 0, and i should use equation Y=Yo+VoT+1/2at^2, where Yo=0, Y=0.83, but i still don't understand how that helps since i dont know Vo and T
  5. Sep 23, 2006 #4
    umm no not quite... Yo would be .83 because that is the initial point at which he jumps and Y=1.85 becuase that is the maximum height he reaches. so once u find Vo from the explanation below you can can find t. dont forget to take into account the time it takes for him to return to the ground though...because if you have done up to this point you have only found about half of the time needed...to get the other time, set Yo=1.85 and Y=.9

    sorry for Vo you can use the equation V^2=Vo^2 +2a(change in y) you should have all these values in order to solve for vo

    P.S sorry if my explanation isnt quite clear. im not a tutor for this site so i dont have tons of practice at explaining things to other people.
    to be honest the trick to doing kinematics problems is to take really small bites at a time. dont try to do it all at once because you will be overwhelmed just try and find anything you can solve for and chances are it will lead you to what you are looking for.
  6. Sep 23, 2006 #5
    hm, that formula for velocity that u gave, isn't that a formula for velocity in one dimensional motion? can it be used here? its not the straight line that the object is moving at..., its a parabola...
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