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FInd the height where two object pass each other.

  1. Sep 8, 2012 #1
    An object is thrown downward with an initial speed of 6 m/s from a height of 42 m above the ground. At the same instant, a second object is propelled vertically up from ground level with a speed of 55 m/s. At what height above the ground will the two objects pass each other?

    The acceleration of gravity is 9.8 m/s²

    2. The attempt at a solution

    Let h = height above the ground where they pass each other.
    42 - h = 6t + 4.95t²
    and h = -55t + 4.95t²
    Adding, 42 = -49t + 4.95t²

    I don't know where to go from here. Should I use Quadratic formula to find the t? I'm confused.
     
  2. jcsd
  3. Sep 8, 2012 #2
    First off, the acceleration should be 4.91t2, not 4.95t2. However, it's much better practice just to leave it as gt2/2.

    Now, allow me to look at the problem in a way that makes more sense to me, and may help you better understand it.
    Let's make an equation for the height of the dropping ball (ball 1):

    [itex]\displaystyle h_1 = 42 - 6t - \frac{gt^2}{2}[/itex]

    Now, an equation for the upwards-traveling ball (ball 2):

    [itex]\displaystyle h_2 = 55t - \frac{gt^2}{2}[/itex]


    The question asks for the height when [itex]\displaystyle h_1=h_2[/itex], or:

    [itex]\displaystyle 42 - 6t - \frac{gt^2}{2}=55t - \frac{gt^2}{2}[/itex]

    I'll let you continue from there. Hope that helps; If you have more trouble just ask.
     
  4. Sep 8, 2012 #3
    How did you get 42 - 6t - gt²/2? Shouldn't it be 42 + 6t - gt²/2? I don't understand.
     
  5. Sep 8, 2012 #4
    The ball is thrown downward at 6 m/s. If you define the starting height to be 42 and the ground to be 0, this means the initial velocity is negative. Another way to put it is that it was thrown at 6m/s toward the ground (toward 0).

    If the height were "42 + 6t - gt²/2" as you say, then that would imply that the ball was thrown upwards at 6m/s (and not downward). The height would first increase to some maximum above 42, then fall past 42 again.
     
    Last edited: Sep 8, 2012
  6. Sep 8, 2012 #5
    ok


    now my try....

    again assume h distance from ground as meeting point at time t

    so h=55t-0.5(9.8)t2(acceleration=-ve as velocity is decreasing)

    and 42-h=6t+0.5(9.8)t2

    solve for h

    i think it gives same result as that of dude14
     
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