A rock is launched at an angle θ = 53.2° above the horizontal from an altitude h = 182 km with an initial speed v0 = [04]___________________

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A rock is launched at an angle θ = 53.2° above the horizontal from an altitude h = 182 km with an initial speed v0 = [04]____________________ km/s. What

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Amity 5 months 2021-08-10T14:31:37+00:00 2 Answers 11 views 0

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    0
    2021-08-10T14:32:58+00:00

    Question is incomplete.

    Complete question is;

    A rock is launched at an angle theta = 53.2 degree above the horizontal from an altitude h = 182 km with an initial speed vo = 1.31 km/s. What is the rock’s speed (m/s) when it reaches an altitude of h/2? (Assume g is a constant 9.8 m/s².)

    Answer:

    Rock’s speed = V = 1870.72 m/s

    Explanation:

    From the question, we are given that;

    h = 182km = 182,000m

    H = h/2 = 91km = 91000m

    Vo = 1.31km/s = 1310m/s

    θ = 53.2°

    g= 9.8 m/s²

    We can find the final speed V at the altitude, H = h/2 by;

    H = h + (VoSinθ)t – 1/2×gt²

    Where,

    t = time taken to reach the height H

    Plugging in the relevant values;

    91000 = 182000 + (1310Sin53.2°)×t – ((1/2) × 9.8 × t²)

    0 = 182000 – 91000 + 1048.96t – 4.9t²

    0 = 91000 + 1048.96t – 4.9t²

    Thus,

    4.90t² – 1048.96t –91000 = 0

    By using quadratic equation, we’ll obtain;

    t = 280.32 seconds and –66.25 seconds

    t cannot be negative and so we pick the positive one.

    So, t = 280.32 s

    Now, V has both horizontal and vertical components which we can calculate;

    The vertical component Vy at t = 280.32s;

    Vy = VoSinθ –gt

    Vy = 1310sin53.2° – 9.8 × 280.32 = –1698.18 m/s

    Now, for the horizontal component, Vx is constant and doesn’t change during the flight of the rock.

    Thus, acceleration is zero. So,

    Vx = VoCosθ = 1310Cos53.2 = 784.72m/s

    Now, the magnitude of the final velocity will be the resultant of Vy and Vx.

    Thus,

    V = √(Vy² + Vx²)

    Plugging in the relevant values,

    V =√(-1698.18² + 784.72²)

    V =√3499600.7908

    V = 1870.72 m/s

    0
    2021-08-10T14:33:04+00:00

    Answer:

    A rock is launched at an angle theta = θ = 53.2° above the horizontal from an altitude h = 182 km with an initial speed v_{0} = 1.31 km/s. What is the rock’s speed (m/s) when it reaches an altitude of h/2? (Assume g is a constant 9.8 m/s^{2})

    The speed of the rocks is 1870.74 m/s

    Explanation:

    The initial potential and kinetic energy of the system is equal to the final kinetic and potential according to the law of conservation of energy.

    Initial (PE + KE) = Final (PE + KE)

    P E = mgh

    K.E = 1/2m

    This implies that;

    mgh_{0} + \frac{1}{2} mv_{0} ^{2}  = mgh_{1} + \frac{1}{2} mv_{1} ^{2}  …………….1

    Where m is the mass;

    h_{0} is the initial height = 182 km = 182 km x 1000 = 182 000 m

    g is acceleration due to gravity = 9.8 m/
    s^{2}

    v_{0} is the initial velocity = 1.31 km/s = 1.31 km/s x 1000 m/km = 1310 m/s

    h_{1} is the final height = h/2 = 182000/2 = 91000 m

    v_{1} is the final velocity;

    Rearranging equation 1 to make  v_{1} the subject formula, we cancel out m;

    gh_{0} + \frac{1}{2} v_{0} ^{2}  = gh_{1} + \frac{1}{2} v_{1} ^{2}

    Substitution into the equation;

    (9.8 x 182000) + (1310)^{2}/2 = (9.8 x 91000) + \frac{v_{1} ^{2} }{2}

    1783600 + 858050 = 891800  +  \frac{v_{1} ^{2} }{2}

    2641650 – 891800 = \frac{v_{1} ^{2} }{2}

       \frac{v_{1} ^{2} }{2} = 1749850

    v_{1}^{2} = 3499700

    v_{1} = \sqrt{3499700}

    v_{1} = 1870.74 m/s


    Therefore the rocks speed is 1870.74 m/s

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