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Physics !!

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1) A stone is dropped from the roof of a building; 2.00 s after that, a second stone is thrown straight down with an initial speed of 25 m/s and the two stones land at the same time.
i) How long did it take the first stone to reach the ground?
ii) How high is the building?
iii) What are the speeds of the two stones just before they hit the ground?

2)A jet fighter is traveling horizontally with a speed of 111 m/s at an altitude of 3.00×102 m, when the pilot accidentally releases an outboard fuel tank.
i) How much time elapses before the tank hits the ground
ii) What is the speed of the tank just before it hits the ground
iii) What is the horizontal distance traveled by the tank.

  • Physics !! -

    first stone
    h=1/2 g t^2
    second stone:
    h=1/2 g (t-2)^2

    set the equations the same, solve for time t it took the first stone
    solve for height of build, h.
    for speed, v^2=2gh

    2. you are given h, solve for t
    h=1/2 g t^2
    speed is a combination of horizontal and vertical speed.
    vertical=sqrt(2gh)
    horizontal given

    speed=sqrt(horizontal^2+vertical^2)

    distance=horizontal speed*time t

  • Physics !! -

    can u help me solve this also ?
    A car from rest and travels for 5.0 s with uniform acceleration of +1.5 m/s2. The driver then applies the brakes, causing a uniform deceleration of -2.0 m/s2. If the brakes are applied for 3.0 s.
    i) How fast is the car going at the end of the braking period?
    ii) How far has it gone?
    iii) Sketch a graph of velocity versus time for the motion.

  • Physics !! -

    Hmmm.

    1) If y is the stone's height at t, and h is the building height, I get

    first stone: y = h - 4.9t^2
    2nd stone: y = h - 25(t-2) - 4.9(t-2)^2

    i: t = 5.63
    the rest is easy...

  • Physics !! -

    For the car problem, just use your standard equations of motion.

    accelerating: v = at = (1.5)(5) = 7.5
    decelerating: v = 7.5 - 2(3) = 1.5

    distance: 1/2 (1.5)(25) + 7.5(3) + 1/2 (-2)(9) = 32.25

  • Physics !! -

    1a. d1 = 0.5g*t^2 = 4.9*2^2 = 19.6 m. =
    Distance fallen by the 1st stone after 2 s.

    V1^2 = Vo1^2 + 2g*d
    V1^2 = 0 + 19.6*19.6 = 384.16
    V1 = 19.6 m/s = Velocity of 1st stone after 2 s.

    d1 = d2-19.6 m.
    Vo1*t + 0.5g*t^2 = Vo2*t+0.5t^2-19.6
    Vo1*t-Vo2*t+4.9t^2-4.9t^2 = -19.6
    19.6t-25t = -19.6
    -5.4t = -19.6
    Tf2 = 3.63 s. = Fall time of 2nd stone.
    Tf1 = 3.63 + 2 = 5.63 s. = Fall time of
    1st stone.

    1b. h = Vo*t + 0.5g*t^2
    h = 0 + 4.9(5.63)^2 = 155.3 m.

    1c. V1^2 = Vo^2 + 2g*h
    V1^2 = 0 + 19.6*155.3 = 3,043.88
    V1 = 55.2 m/s.

    V2^2 = 25^2 + 19.6*155.3 = 3668.88
    V2 = 60.6 m/s.

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