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2 dimensional motion problem

  1. Jan 23, 2013 #1
    The problem:

    A ball is thrown from the top of a building at an angle of 51° to the horizontal with an initial speed of 24m/s. The ball is thrown at a height of 59 meters from the ground and hits the ground at 88.5208 meters from the base of the building.

    What is the speed of the ball just before it strikes the ground?


    Relevant equations:

    x(t) = x(initial) + v(initial)t + 0.5at^2

    v(t) = v(initial) + at


    My attempt at solution:

    I initially wanted to find the time at which the ball hits the ground. Using the x as a function of time equation, I surmised I needed to also solve for v(initial in the x direction) to help me solve for t. Using trig, I defined v(initial in the x direction) as 24cos51. I defined other terms in this equation:

    x(t) = 88.5208 meters
    v(initial)t = 24cos51(t)
    0.5at^2 = -4.9t^2

    Using the quadratic formula, t = 1.5 seconds.

    I then input 1.5 for t in the equation: v(t) = v(initial) + at. My guess is that if I have the total time the ball travelled, I can determine its velocity at the point of impact.

    v(t) = v(initial) + at
    v(initial) = 24 m/s
    a=-9.8 (1.5)

    yielding the velocity of the ball to equal -8.9m/s.

    I'm not entirely sure I went about solving this problem in a way that is correct. Any suggestions in helping me understand the process of solving this problem are greatly appreciated. Thank you
     
  2. jcsd
  3. Jan 23, 2013 #2

    jedishrfu

    Staff: Mentor

    You want to work with the y direction first to get the time until the ball hits the ground and then use it to find x distance traveled.
     
  4. Jan 24, 2013 #3
    Thanks for your advice. I tried it again, this time using the y direction to solve for the total time the ball is in the air. I guess that makes more sense, since the ball starts at 59 meters in the y direction? Therefore it must travel in the negative y direction in the first place?

    In doing so, I let 0=y(initial) +v(initial y)t +0.5at^2

    y(initial) = 59 meters
    v(initial y)t = 24sin51(t)
    0.5at^ = -4.9t^2

    Quadratic formula yield t = 5.86 seconds.

    Using d=vt:
    d= 88.5208
    v=?
    t=5.86 seconds

    solve for v. v=-15.1m/s
     
  5. Jan 24, 2013 #4

    jedishrfu

    Staff: Mentor

    looks good up to the point of the d=vt you have vx its v iniitial * cos(51) and you need to find vy at 5.86 secs then add them vectorially and then find the length of that vector to get the speed.
     
  6. Jan 24, 2013 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Are you required to do it that way? It seems to me that, because they are asking for speed, rather than velocity, it would be much easier to use "conservation of energy".
     
  7. Jan 24, 2013 #6

    jedishrfu

    Staff: Mentor

    True, I was just following the OP's line of thought.
     
  8. Jan 24, 2013 #7

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    There are lots of ways to solve this one but I think I agree that using CoE is the best way.

    The ball has PE and KE at the start totaling...

    mgh + 0.5mVi2

    On landing it has zero PE and KE totaling..

    0.5mVf2

    Equate

    mgh + 0.5mVi2 = 0.5mVf2

    Mass cancels. Rearrange to give..

    Vf2 = 2gh + Vi2

    Vf = SQRT(2gh + Vi2)

    or something like that.

    Had they asked for the angle whith which it hits the ground then I'd probably calculate the horizontal and vertical components at impact using standard equations of motion.
     
  9. Jan 24, 2013 #8
    I think I did it right this time. I went back to this equation:

    v(y) = v(initial position in y direction) + at

    v(initial position in y direction) = 24sin51
    at = -9.8(5.6 sec) = -57.428

    v(y) = -38.78m/s

    Not required. I'm in intro level physics and we've been taught from 1 and 2 dimension motion up thru Newton's 2nd law. But if there's a more direct approach to solving these types of problems..I can't wait :P

    I tried it this way, too and got -24.09m/s.

    My two answers don't match up. (The other one being -38.78m/s)

    Sorry guys, I'm definitely on the struggle bus on this one.
     
  10. Jan 24, 2013 #9

    jedishrfu

    Staff: Mentor

    I got 41.6 using the CoE way not -24.09 maybe check what you did.

    I used this expression in google to get it calculated: sqrt(2*9.8*59 + 24*24)
     
  11. Jan 24, 2013 #10
    Set v = [itex]\sqrt{V(x)^2 + V(y)^2}[/itex]

    V(x) = V(ox) + at
    V(ox) = 24cos51
    at = 0 (because no acceleration in the x direction)
    Vx^2 = 228.121

    V(y) = V(oy) +at
    V(oy) = 24sin51
    at = -57.428
    Vy^2 = 1503.616

    Add 'em together and take the square root...41.61m/s
     
  12. Jan 24, 2013 #11
    Oh yay we got the same answer :)

    Thanks a ton. I'm not very good at these types of problems yet...but I will keep practicing.
     
  13. Jan 24, 2013 #12

    jedishrfu

    Staff: Mentor

    Yes, please practice. We need you here at PF to help future students trying to solve these kinds of problems...
     
  14. Jan 25, 2013 #13

    CWatters

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    Science Advisor
    Homework Helper

    It's quite hard for a square root to give a negative answer :-)

    I got 41.6m/s
     
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