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28 December, 16:02

A mortar shell is fired with a muzzle speed of 300 ft/sec. Find the angle of elevation of the mortar if the shell strikes a target located 1500 ft away. Round your answer to 2 decimal places.

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  1. 28 December, 18:23
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    Angle of elevation = 4.37°

    Step-by-step explanation:

    Velocity of the mortar shell = 300ft/sec

    Distance covered = 1500ft

    Range = U²sin2tita/g

    But range = distance

    Range = 1500ft

    U = velocity

    g = acceleration due to gravity

    Tita = angle of projection of elevation

    1500 = 300²sin (2tita) / 9.87

    (1500*9.87) / 90000 = sin2tita

    0.1645 = sin2tita

    Sin^-1 0.1645 = 2tita

    9.4682 = 2tita

    total = 9.4682/2

    tita = 4.3741°

    To two decimal place = 4.37°
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