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3 December, 10:19

A gun has muzzle speed 144 m/s. find two angles of elevation that can be used to hit a target 780 m away. (enter your answers as a comma-separated list. round your answers to one decimal place. use g â 9.8 m/s2.)

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  1. 3 December, 10:28
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    The muzzle speed is 144 m/s.

    Let x = the angle of elevation.

    Then

    u = 144 cos (x), the horizontal velocity

    v = 144 sin (x), the vertical launch velocity

    Assume no air resistance, and g = 9.8 m/s².

    To hit the target 780 m away, the time of flight is

    t = (780 m) / [144 cos (x) ] = 5.4167 sec (x) s

    In terms of the vertical velocity,

    (144 sin (x) m/s) * (5.4167 sec (x) s) + (1/2) * (-9.8 m/s²) * (5.4167 sec (x) m/s) ² = (780 m)

    780 tan (x) - 143.769 sec²x = 780

    tan (x) - 0.1843 (1 + tan²x) = 1

    0.1843 tan² (x) - tan (x) + 1.1843 = 0

    tan²x - 5.4259 tan (x) + 6.4259 = 0

    Solve with the quadratic formula.

    tan (x) = 0.5[5.4259 + / - √ (3.737) ] = 3.6795 or 1.7464

    x = tan⁻¹ 3.6795 = 74.8°

    or

    x = tan⁻¹ 1.7464 = 60.2°

    Answer: (74.8°, 60.2°)
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