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6 July, 20:01

The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give the values r = 32700 ft, r¨ = 85 ft/sec2, and θ˙ = 0.019 rad/sec. Calculate the magnitudes of the velocity and acceleration of the rocket at this position.

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  1. 6 July, 20:16
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    velocity = 1527.52 ft/s

    Acceleration = 80.13 ft/s²

    Explanation:

    We are given;

    Radius of rotation; r = 32,700 ft

    Radial acceleration; a_r = r¨ = 85 ft/s²

    Angular velocity; ω = θ˙˙ = 0.019 rad/s

    Also, angle θ reaches 66°

    So, velocity of the rocket for the given position will be;

    v = rθ˙˙/cos θ

    so, v = 32700 * 0.019 / cos 66

    v = 1527.52 ft/s

    Acceleration is given by the formula;

    a = a_r/sinθ

    For the given position,

    a_r = r¨ - r (θ˙˙) ²

    Thus,

    a = (r¨ - r (θ˙˙) ²) / sinθ

    Plugging in the relevant values, we obtain;

    a = (85 - 32700 (0.019) ²) / sin66

    a = (85 - 11.8047) / 0.9135

    a = 80.13 ft/s²
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