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19 July, 09:07

During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61 km/s at an initial inclination of 81.9° to the horizontal. The acceleration of gravity is 9.8 m/s^2. How far away did the shell hit? Answer in units of km How long was it in the air? Answer in units of s.

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  1. 19 July, 11:10
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    The shell hit at a distance of 1.9 x 10² km

    The time of flight of the shell was 5.3 x 10² s

    Explanation:

    The position of the shell is given by the vector "r":

    r = (x0 + v0 * t * cos α; y0 + v0 * t * sin α + 1/2 g t²)

    where:

    x0 = initial horizontal position

    v0 = magnitude of the initial velocity

    t = time

    α = launching angle

    y0 = initial vertical position

    g = acceleration of gravity

    When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

    Then:

    ry = 0 = y0 + v0 * t * sin α + 1/2 g t²

    Since the gun is at the center of our system of reference, y0 and x0 = 0

    0 = t (v0 sin α + 1/2 g t)

    t = 0 is discarded as solution

    v0 sin α + 1/2 g t = 0

    t = - 2v0 sin α / g

    t = (-2 * 2610 m/s * sin 81.9°) / (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

    Then, the distance at which the shell hit is:

    Distance = Module of r = (x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α

    Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
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