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Problem with gravitational acceleration

  1. Mar 1, 2006 #1
    A person standing at the base of a building throws a stone straight up at 40 m/s. If the building is 35 m high, calculate:

    a) The maximum height to which the stone will rise
    b) How long it would take for the stone to land on the roof
    c) The velocity at which the stone would hit the roof

    I was only able to calculate a (I got -81.6m). It took 4.08 s for the stone to get to that height. I have no clue how to get b and c. If the distance from the max height to the roof is 46.6 m, the acceleration is 9.81 m/s2, and V1 is 0m/s, I can't find a way to calculate V2 and the time.
  2. jcsd
  3. Mar 1, 2006 #2
    Hey dude, no worries. Remember that there are 3 kinematics equations in one dimension, two involving time and one without. What you are given is the h (delta y) and the initial velocity, V0. Because there is no angle (the ball is thrown straight upwards) everything can be found with those three simple equations. Try and dig them up from your notes and give it another shot and let us know how it went.
  4. Mar 2, 2006 #3
    Keep in mind the physical significance of the answer you got for part a. You say it's negative, which means you set up your axis pointing from the sky down to the ground. If that's not the case, you should re-evaluate how you did it.

    For the sake of completeness and so we can see the whole picture, I'll walk you through all the steps which are critical to be able to tackle problems such as these.

    First set up a coordinate system. The most logical thing is to set the origin on the ground, with the positive axis pointing upward. This is important in this problem as it will tell us which sign to use for the acceleration due to gravity. The ag is pointing towards the ground, in the opposite direction of our axis, so we'll take ag = -g.
    This will give us a positive distance, since any point above the ground is on the positive axis.

    Now the most important "trick" to doing a problem like this: Do it in segments! In this case, we'll do it in 2 parts.
    1) The stone goes from the ground up to its max point in the air.
    2) The stone falls from the max point down to the roof.

    1) As you already discovered, we can solve the distance as follows:
    vf is zero, since the stone stops momentarily at the top of its path. Examining the available equations, we see that we can solve for the time it takes to reach the top of the path.
    [tex]v_f = v_0 + a_g t = v_0 - gt \approx 40 \mbox{m/s} - 9.8 \mbox{m/s^2} t[/tex]
    Now solve for t.

    With t, we can now find the distance.
    [tex]d = v_0 t + \frac{1}{2}a_g t^2 = v_0 t - \frac{1}{2}gt^2[/tex]
    With our coordinate system, d will be positive.

    2) Now, we start treating this as a mostly-new problem to solve. As you discovered, the stone is now about 42 meters above the top of the building.

    What do we know to start out with:
    v0 is zero, for the same reason our vf was zero in the previous part (the stone stops momentarily before it turns and begins falling).

    ag is still -g.

    It's 42m above the building -> This means it has to fall 42m. So the d we'll use should be -42m (negative due to our choice of coordinate system earlier).

    Part (b) asks how long it will take for the stone to land on the roof. We already solved for the time it takes to reach the top of its arc, let's call that t1.

    Now figure out how long it takes the stone to travel a d=-42m, and call this t2.

    So the total time is just t1+t2.

    Part (c) asks what its velocity will be when it hits the roof. Again, we're going to work this in our second segment, using v0=0, and our t will be the same t2 we used to find part (b).
    There are a couple of ways to solve this, but the easiest is probably
    [tex]v_f = v_0 + a_g t = v_0 - g t_2 = 0 - g t_2[/tex]

    I've gone into considerable detail in trying to step through this problem, which others on this forum might not agree with. However, all too often students cludge their way through these early, introductory physics problems and slide through without gaining a real understanding for the "proper" way to approach the problems.

    This ultimately hurts them later on.

    Alain12345 - if any of this (especially the reasoning, or why I chose certain things to be certain ways) doesn't quite make sense, please ask! There are a wealth of people here who are happy to explain things. :)
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