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Physics- falling objects, parabola shape

  1. Nov 4, 2003 #1
    Hey!

    I know that the horizonal and horizontal motion are unrelated, and the velocity remains the same (if there is no air resistance, friction etc)- this correlates with the whole 'shooting a gun and dropping a bullet at the same time, which one will hit the ground first? They will fall at the same time.' However, I don't know how to find the different parts of a problem.

    For example:

    A stone is thrown horizontally at a speed of 10 m/s from the top of the cliff 78.4 cm high. How long does it take the stone to reach the bottom of the cliff? How far from the base of the cliff does the stone strike the ground?

    How also would I find how fast a ball is going if it ended up going X far off a cliff X high? Or, how would I find how high the cliff is if I have the other two pieces of data?

    Thanks!

    I don't need a specific answer to the example above, I just need to figure out how to go about getting the answer for each given type.

    I know d=1/2gt squared...

    I know I need to switch around a formula! Which one?
    :frown:

    (yes, I posted this in homework help as well...)
     
  2. jcsd
  3. Nov 4, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You can, as you said, separate horizontal and vertical components.
    In particular, the only force is vertical: gravity= -mg. If you want to write it in terms of a vector: F= <0, -mg>

    Since there is no force horizontally, there is no acceleration horizontally (F= ma, F= 0 so ma= 0=> a= 0). The acceleration vertically is (-mg= ma so a= -g) -g. The acceleration vector is <0, -g>.

    The velocity is, of course, initial velocity plus change in velocity= initial velocity + at. Horizontally, the initial velocity is 10, a= 0 so v= 10 for all times t. Vertically, the initial velocity is 0 so v= 0- gt. The velocity vector is <10, -gt>

    The position is the initial position plus the change in position. For constant velocity the change in position is velocity times time.
    Since horizontal velocity is 10, and we choose the 0 point to be at the top of the cliff, the horizontal position is 0+ 10t= 10t.

    For variable velocity, we have to integrate. Since the vertical velocity is -gt, the position is: initial position + (-g/2)t2. It would be reasonable to take either the top of the cliff or the bottom of the cliff.
    If we take the bottom of the cliff as 0, then the initial position is +78.4 m so the vertical position at time t is 78.4- (g/2)t2. The position vector is (10t, 78.4- (g/2)t2
    If we take the top of the cliff, as 0, then the initial position is 0 so the vertical position at time t is 0- (g/2)t2. The position vector is (10t, -(g/2)t2.

    To determine "how far from the base of the cliff does the stone hit the ground", we have to set the vertical position to the coordinate of the "ground". If we chose the bottom of the cliff as 0, then the ground is at 0. Vertical position 78.4-(g/2)t2= 0. If we chose the top of the cliff as 0, then the ground is at
    -78.4. Vertical position -(g/2)t2= -78.4. Those are, in fact,the same equation and have the same solution, t. Use that t in "horizontal position= 10t" to find the vertical position, the distance from the cliff, when the stone hits the ground.
     
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