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Thread: Physics...

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    Physics...

    If I could get some help with these four, I could use the same techniques to solve the remainder of the problems...

    Motion In One Dimension:
    1. <LI>A ball is thrown from a building straight down and a second ball is thrown from the same building straight up in the air with the same initial speed. Compare the final speeds of the balls as they reach the ground.
      <LI>A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground six seconds later. What was the rocket's acceleration?
      <LI>Divers compete by diving into a 3.0m deep pool from a platform 10m above the water. What is the magnitude of the minimum acceleration in the water needed to keep a diver from hitting the bottom of the pool? The acceleration is constant.
      <LI>A rocket-car is launched along a track at t = 0s. It moves with constant acceleration 2.0 m/s^2. At t = 2.0s, a second car is launched with constant acceleration 8.0 m/s^2. At what time does the second car catch up with the first?
    Thank you,

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    Re: Physics...

    Hi Jordan,

    May I be the first to welcome you to the lounge.
    Obviously, I am unable to answer your questions but I will help you find a suitable userpic! <img src=/S/smile.gif border=0 alt=smile width=15 height=15>

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    Re: Physics...

    If I remember my physics correctly (and it's highly possible that I do not - it has been a long time)..
    1) <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">They will hit the ground at the same speed. Since the ball going into the air had the same speed going up, it will have the same speed when it gets to the "launching pad", matching the first balls initial speed as it descends toward the ground.</span hide>

    Just found this to support my answer wow! I remembered

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    Re: Physics...

    And here is Q4.
    Attached Images Attached Images
    Regards
    John



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    Re: Physics...

    I am not sure that Q3 has enough info. The "minimum" is a little misnomer. The minimum is when the diver does not dive but just falls from the platform. If the diver actually dives, the minimum required to stop him or her will be higher than the mimimum if they fall. Thus the number calculated will not stop a diver from hitting the bottom, it will only stop someone who falls from the platform.

    There are 2 required equations:

    V = V0 + a * t

    X = X0 + V0 * t + a / 2 * t^2

    The minimum acceleration required will be when the diver does not dive, but just falls from the top platform. At this point the distance (x0) is 13 m (3 m from bottom to top of pool and 10 m from top of pool to platform). The starting velocity (V0) is zero and the distance traveled until the water is reached is 10 m. The only acceleration is that due to gravity and this is -9.8 m/s/s. Therefore:

    3 = 13 + 0 * t + (-9.8)/2 * t^2

    Rearranging:
    -10 = -4.9 * t^2
    t = sqrt(10/4.9) = 1.429 sec

    With the time we can calculate the velocity of the swimmer as it enters the water:

    V = V0 + a * t
    V = 0 + (-9.8) * 1.429 = - 14.0 m/s

    We can then calculate the maximum time we have. This is the time to hit the bottom of the pool with no water in it.

    X = X0 + V0 * t + a/2 * t^2

    0 = 3 + (-14) * t + (-9.8)/2 * t^2

    This can be solved with the quadratic equation to yield t = 0.200 sec. If we want the velocity to be zero at this time we can solve:

    V = V0 + a * t

    0 = -14 + a * (0.200)
    a = 14.0 / 0.200 = 70.0 m/s/s

    Since this is the total acceleration, the acceletation due to the pool is:
    70.0 - (-9.8) = 79.8 m/s/s

    If the diver does not fall from the platform, but dives upward and then comes downward the acceleration will need to be larger since the time to reach the bottom will be smaller with the increased downward velocity.

    Steve

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    Re: Physics...

    And here is an answer for Q2.

    added later Oops..I just noticed that the final line of the non-metric answer says 10, when it should say 18.
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    Regards
    John



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    Re: Physics...

    Q1. A simple way to look at this is through the conservation of (potential + kinetic) energy. Since the initial states are the same and the final states of potential energy correspond to ground level, the final states of kinetic energy (and hence speed) must be the same.

    Alan

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    Re: Physics...

    Thank you so much! I really appreciate the help.

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    Re: Physics...

    I checked with my father who is a mechanical engineer. His answers are as follows:

    1. When the balls hit the ground both will have zero velocity.

    2. Depends on the state of the moon and its gravitational pull. .

    3. The diver needs to decelerate not accelerate to avoid hitting the bottom. Typically the co-efficient of friction between the diver and the water is such that the constants of deceleration are not constant. The divers body is subjected to changing dynamic forces which affect the body's shape thus changing the physics as the body travels through the water. That being said the diver will still have a head ache, particularly if the diver comes to rest at the bottom of the pool and stays there. Nitrogen can cause the bends, but then everything is in a straight line so the bends do not apply.

    4. Just after the Tortoise passes the Hare.

    Hope this helps!
    ...

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    Re: Physics...

    Thanks again for your help. I have the solution to Q3...

    This is what we know...
    s(vertical in air) = 10m
    s(vertical in pool) = 3m
    g = -9.81m/s^2
    Vf (in pool) = 0m/s
    Vi (in pool) = Vf (in air)
    Vi (in air) = 0m/s because you're trying to find the least possible acceleration

    We want to find the acceleration (in the pool)

    Using the equation Vf^2 = Vi^2 + 2as, we can find his velocity as he hits the water. Use gravity for a b/c this is a free-fall.
    Vf(air) = Vi(pool) = sqrt[ 2 * (-9.81m/s^2) * (10m) ]
    Vf(air) = -14m/s

    We can use the same equation to find acceleration, but this time we have to use s(in pool).
    a = (Vf^2 - Vi^2)/(2s)
    a = (0 + (14m/s)^2)/(2 * 3m)
    a = 32.7m/s^2

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