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Falling objects help!

  1. Oct 15, 2004 #1
    Hey guys, just wanted to thank everyone for some help (mainly from other posts) that have saved me quite a headache in the past. Now I am having some major issues with falling objects and formulas and such... I missed about 4 days of class and am wayyyyyyy behind. So, with that said here we go...

    A stone is thrown vertically upward with a speed of 13.5 m/s from the edge of a cliff 66.0 m high. What is the speed of the stone right before hitting the ground? What is the total distance the stone traveled? and lastly, How much later does it reach the bottom of the cliff?

    I tried to use the formula:

    Y_f=y_i +V_i(t)+1/2gt^2

    I come up with what seems like random awnsers and it confuses me :frown: Any assistance would be greatly appreciated because this problem gives me one heck of a headache.

    Thanks much,
    Spd :cry:
     
  2. jcsd
  3. Oct 15, 2004 #2

    quasar987

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    The trick is to separate the problems in 2 parts: 1) the ascension of the stone and 2) its falling.

    You throw your stone upward vertically at 13.5 m/s. First let's define a convenient coorinate system. We will take the positive y axe as upward. During its trip upwards, there is a force [itex]F=-mg[/itex] acting on it. Newton's second law of motion says the force acting on an object equals its acceleration. So we have, [itex]-mg = ma_y \Leftrightarrow a_y = -g = -9.80 \frac{m}{s^2}[/itex]. The acceleration of an object is the rate as which its speed changes. In mathematical form, [itex]a_y=\frac{\Delta v_y}{\Delta t} \Leftrightarrow \Delta t = \frac{\Delta v_y}{-a_y} \ (*)[/itex]. And this is when you must use a bit of your physical intuition: at its maximum height, the speed of the stone is zero. So it starts at 13.5 m/s and end up at 0 m/s. This means that for the first part of the problem (as we have astuciously defined it), [itex]\Delta v_y = v_y_f - v_y_i = 0 - 13.5 = -13.5 \frac{m}{s}[/itex] We can use this result in equation [itex](*)[/itex] to find the TIME that it took for our speed to drop from 13.5 m/s to 0. And now you have all the informations you need to use your equation [itex]y_f=y_i+v_y_it + \frac{1}{2}gt^2[/itex].

    And for the second part of the problem, all of your final values from the first part become your initial value for the second part (because the second part is just the continuation IN TIME of the first part).

    I think all you missed is the subtelty of dividing the problem in half, so you shouldn't have difficulty solving the second part.
     
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