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Projectile motion (time cut in half)

  1. Mar 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Two buildings A & B are each of height H as measured from the ground & are located a distance L apart. A tennis ball is shot horizontally with a velocity vo from building A such that it just hits the bottom of building B before hitting the ground. If building B is moved to half the distance (1/2)(L) & the ball is launched with the same horizontal velocity, where on building B does it hit the side, measured from the ground?

    2. Relevant equations
    I think they are:
    x-xo = (vocos(theta))t
    y-yo = (vosin(theta))t - 1/2(g)t2

    3. The attempt at a solution
    Where the ball would be at L/2 = x/2 & results in t/2 because x is proportional to time in the equation.
    So apply half the time to the y axis equation, but since it's 2 terms with different degrees of t (one is t1, the other t2) I'm not sure what I can assert about how this effects y.
     
  2. jcsd
  3. Mar 24, 2017 #2
    I just realized that the y axis won't use the equation I posted, the y axis is a one dimensional motion... I'll try this again
     
  4. Mar 24, 2017 #3
    Okay, is this correct:
    We can deduce that it'll be half the time from my reasoning in the first post.
    the y axis equation we need to use is: Δy = vot + ½at2
    since vo = 0, Δy = ½at2,
    if we apply t/2 we get Δy = ½a(½t)2 =(¼)½at2, so we'll end up with ¼*Δy, giving us 3/4H from the ground.

    Is this all sound reasoning?
     
  5. Mar 24, 2017 #4

    cnh1995

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    Looks good!
     
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