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Homework in college

  1. Sep 25, 2003 #1
    Astone is thrown up vertically upward from the edge of a building 19.6 meters high with an inital velocity of 14.7 m/sec. The stone just misses the building on the way down and strikes the street below.
    A) time to reach max height
    b) total time of flight
    C) height above the building
    d) the velocity of the stone before it hits the ground

    I think I am on the right track could anybody tell me if they are right?
    A) 0 - 14.4m/se / -9.80 Where the time to the max height 1.5 sec
    c) Ymax = Vo(t) + .5(-g)(t squared) the formula use
    Ymax = 22.05 -11.03
    Ymax is 11.0 meters above the building
    D) Height above ground = building height + stone max height
    19.6m + 11.0m = 30.6 m
    B) 2h/g = t squared Which Tdown is = 2.5 sec and Tup is 1.5 secand get 4.0 secs.
    Should this be more time becaues the stone would have been in flight for 3 sec. When it becomes = with the build on the way down before accel more?
    E) Vf = g (Tdown) Vf = -9.80 (2.5)
    Vf is = to -24.5 m/sec

    Thanks you

  2. jcsd
  3. Sep 25, 2003 #2


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    a) is correct. The stone will have speed upward after t seconds of
    14.7- 9.8t and will continue upwards until its speed is 0:
    14.7- 9.8 t= 0 so t= 14.7/9.8= 1.5 sec. Since it is going downward after that, that is the highest point.

    For (c)I get 11.025 meters but I won't quibble!

    On (d) you've lost me completely. Did you drop (d) and miswrite (e) for (d)? (d) asks for speed when it hits the ground but you are calculating maximum height above ground.(Of course. Just add 19.6 to the answer to c.)

    e) speed just as the stone hits the ground ("speed before it hits the ground" is ambiguous. Perhaps "JUST before it hits the ground"?):

    You know, since you've already used it that
    v(t)= 14.7- 9.8t and that h(t)= 19.6+ 14.7t- 4.9t2.

    The stone will hit the ground when h(t)= 0 so solve
    19.6+ 14.7t- 4.9t2=0 and put that value of t into
    v= 14.7- 9.8 t. (There will be two times: one positive and one negative. Decide which to use!)
    I get -63.7 m/s (the stone is going down at 63.7 m/s as it hits the ground.
  4. Sep 25, 2003 #3
    a) v = u + at
    0 = 14.7 - 9.81(t)
    t = 1.50s

    S = ut + 1/2at^2
    S = 14.7(1.5) - 0.5(9.81)(1.5^2)
    = 22.05 - 11.03625
    = 11.01375m

    b) v^2 = u^2 + 2as
    = 0 + 2(9.81)(11.01375+19.6)
    v = 24.5ms-1

    v = u + at
    24.5 = 0 + 9.81(t)
    t = 2.498s

    Total time = 4.498s

    c) Height = 11.01375 + 19.6 = 30.61375m

    d) v = 24.5ms-1
  5. Sep 25, 2003 #4
    Damn, HallsofIvy you beat me to the answer :wink:

    BTW, are you sure of the 63.7ms-1 figure?

    Unless I am reading the question wrong the stone is effectively 'dropped' with a zero initial velocity at its apex. The drop height is about 30m and without checking too much, common sense says 63.7 is a little high.
  6. Sep 25, 2003 #5


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    You're right. By "conservation of energy", if it falls from 30 feet it should reach a speed of 25.5 m/s.
  7. Sep 26, 2003 #6


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    I SQUARED t in the velocity calculation!

    The correct speed as the stone hits the ground is -24.7 m/s.

    You could also do this using "conservation of energy" based on the "initial state" information given:
    Initially, the stone was at height 19.6 meters and so has potential energy -mgh= -m(-9.8)(19.6)= 198.02m Joules. It has speed 14.7 m/s and so kinetic energy (1/2)mv2= 108.045m Joules, making the total energy 306.065m Joules.

    When it hits the ground, its height is 0 so it has 0 potential energy- all of its energy must be kinetic energy:
    (1/2)m v2= 306.065m Joules. We can cancel the mass, m, mulitply by 2 and have v2= 612.13.
    v is the square root of that: -24.7 m/s. (The negative square root because we know it is going downward.)
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