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Object Thrown from building help

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A boy throws a stone from the top of the building that is 45 m high. It lands 64 m away. Find the following:

    a) time of flight
    b) initial speed
    c) speed and angle with respect to the horizontal of the velocity vector at impact

    2a) throwing the stone with more force (1.5 times harder) multiplies the time it takes to reach the ground by?

    2b) The horizontal component of the velocity is multiplied by what factor?

    2c)how many times further does the stone land away from the building

    2. Relevant equations

    d=1/2gt^2
    Xf=Xi+Vot


    3. The attempt at a solution

    a) I figured out that since it take 3.03 seconds to drop 45m then it must also take 3.03 for it to land when being thrown horizontally. solved using d=1/2gt^2

    b) knowing time: 64=Vot => 64/3.03=Vo ----> Vo = 21.13m/s

    everything else I can't seem to figure out
     
  2. jcsd
  3. Sep 22, 2013 #2

    gneill

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    Staff: Mentor

    For part (c), can you determine the vertical component of the velocity of the stone at impact? How about the horizontal component?
     
  4. Sep 22, 2013 #3
    The total velocity would be sqrt(ViSinθ^2 + ViCosθ^2) so would it be ViSinθ = -(sqrt(ViCosθ^2))??
     
  5. Sep 22, 2013 #4

    gneill

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    Staff: Mentor

    Well, you don't know the angle of impact (yet), so that wouldn't be helpful. And the magnitude of the final velocity won't be equal to the initial velocity Vi, since the stone will be gaining energy as it falls from its initial elevation to the ground below.

    So, let me offer another viewpoint. You know that the horizontal and vertical components of projectile motion are independent (at least until the flying object hits something!). Consider each component separately. Start with the vertical component. How fast will the stone be traveling vertically having fallen 45m from rest?
     
  6. Sep 22, 2013 #5
    it would be falling with V = -gt. d=1/2gt^2 would get me 3.03 seconds to fall from rest. so -gt = -29.71m/s that is the vertical component itself? considering the horizontal component always stays the same because no force acts on it to provide any accel.
     
  7. Sep 22, 2013 #6

    gneill

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    Right. So now you know both the vertical and horizontal components of the velocity at impact (assuming you can calculate the horizontal component given your distance and time).
    What can you determine from that?
     
  8. Sep 22, 2013 #7
    that final velocity of impact is sqrt(29.71^2 + 21.12^2) will give me the velocity right before it hits the ground.
    ^^ I think that is wrong

    the only other thing I can think of is that the components can be further broken down to i + j form and then placed into each other to get one final equation?
     
  9. Sep 22, 2013 #8

    gneill

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    The components are already in "i + j form", since one is vertical and the other horizontal; they are at right angles to each other, and directed along coordinate axes. So essentially they are orthogonal vectors.

    So your final velocity magnitude is okay. What about the angle?
     
  10. Sep 22, 2013 #9
    if they are in I + J form already then tan^-1(j/i) gives me the angle measure which is -54.6 degrees?
     
  11. Sep 22, 2013 #10

    gneill

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    Sure. Looks good.
     
  12. Sep 22, 2013 #11
    Got it!! took a second look at it and became so obvious why. thank you!
     
  13. Sep 22, 2013 #12

    gneill

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    Staff: Mentor

    It's always a good feeling when what looked difficult suddenly becomes obvious :smile:

    Cheers!
     
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