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16 August, 06:02

a missile is moving 1350 m/s at a 25.0 deg angle. it needs to hit a target 23,500 m away in 55.0 deg direction in 10.20 s. what is the magnitude of its final velocity?

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  1. 16 August, 06:14
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    In this question we have given

    velocity of missile=1350m/s

    angle at which missile is moving=25degree

    distance between missile and targets=23500m

    angle between target and missile=55degree

    time=10.2s

    To find the final velocity of missile we will first find the acceleration required

    Let x be the horizontal component of distance

    x - vertical component of distance

    t-time

    ax - horizontal component of acceleration

    ay-Vertical component of acceleration

    Vx-horizontal component of velocity

    Vy-Vertical component of velocity

    horizontally: x = Vx*t + ½*ax*t²

    23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s) ²

    ax = 19.2 m/s²

    V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

    similarly vertically:

    y = Vy*t + ½*ay*t²

    23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s) ²

    ay = 258 m/s²

    V'y = Vy + ay*t = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

    Therefore

    V = √ (V'x² + V'y²) = 3504 m/s

    therefore magnitude of final velocity of missile=3504m/s
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