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Calculating velocity, time, and Instantaneous speed using kinematic equations

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    An inquisitive physics student and mountain climber climbs a 52.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.10 m/s.


    (a) How long after release of the first stone do the two stones hit the water?


    (b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?

    (c) What is the speed of each stone at the instant the two stones hit the water?
    first stone m/s
    second stone m/s

    2. Relevant equations

    vf^2=vi^2+2a(xf-xi)
    xf=xi+vi t+1/2at^2

    3. The attempt at a solution

    I attempted to find the T

    xi=52m
    xf=0?
    vi=2.10m/s
    a=-9.81m/s^2
    vf= (by the first equation) = 32m/s

    I tried plugging in the result in the second equation, but it's starting not to make sense.
    I was told that I have to assume the stones to have been thrown vertically but with an angle, because it wouldn't make sense otherwise, but I am having difficulty getting what that angle is, and this assignment does not include the trig functions.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 16, 2012 #2
    If the stones are thrown vertically downward, you can think of the angle being 90° (the angle between the positive x-axis and the negative y-axis). The horizontal component of the velocity (magnitude*cos90) = 0 and the vertical component of the velocity (magnitude*sin90) is just the original magnitude. In that aspect, this can be thought of as a 1-dimensional problem.

    That would work. The distances are relative to each other; you are only interested in the difference between the distances. (The second equation can be rewritten as xf-xi=vi t+1/2at^2). You could have also, for example, set xi=0 and xf=-52.

    Also, don't forget the equation vf=vi+at :wink:

    Edit: If you were told that the angle wasn't 90° and you weren't given the angle, then you don't have enough information to solve the problem. Although, 90° seems plausible since the stones can be thrown vertically if the student reaches over the ledge.
     
    Last edited: Feb 16, 2012
  4. Feb 16, 2012 #3

    So I cannot consider the acceleration as -9.81 because this isn't a free falling, right?
     
  5. Feb 16, 2012 #4
    The acceleration of gravity on Earth is always -9.81m/s^2. An object speeding at 3000m/s towards the ground will feel the same -9.81m/s^2 acceleration as an object that has just been dropped. In reality, an object going at 3000m/s might be slowing down if the air resistance is large enough, but the gravitational acceleration is still -9.81m/s^2 (assuming it's close to the Earth's surface).
     
  6. Feb 16, 2012 #5
    But if the vf=32 m/s then using the second equation

    32=2.10+ (-9.81)t
    the t = -3.04 which isn't possible. I am not sure where I went wrong.
     
  7. Feb 16, 2012 #6
    Velocity is a vector. Since the velocity is downwards, it is negative. When you took the square root to solve for the final velocity, you wrote + instead of +/-
     
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