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HELP! A cannon tilted up at a 35 ∘ angle fires a cannon ball at 81 m/s from atop a 18-m-high fortress wall?

What is the ball's impact speed on the ground below? What is the ball's impact speed on the ground below? PLEASE HELP ASAP!!

2 Answers

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  • 3 months ago
    Favorite Answer

    Vx = 81 cos 35 = 66.4 m/s

    Vy = 81 sin 35 = 46.5 m/s

    from the 18 m, how high does it go vertically

    h = v²/2g = 46.5²/2•9.8 = 110.1 m

    add 18 to that, total height = 128.1 m

    speed after falling 128.1 m

    v = √(2gh) = √(2•128.1•9.8) = 50.1 m/s

    speed is vector sum horiz. and vert. speeds

    v = √(66.4² + 50.1²) = 83.2 m/s

  • 3 months ago

    Height h(t) = ho +Vyo*t - 4.9t² =  18 + 81sin35 *t - 4.9*t² = 0 at t = 9.854s

    Vyf = Vyo - g*t = 81*sin35 - 9.8*9.854s = -50.11m/s

    Vxf = Vxo = 81*cos35 = 66.35m/s

    50.11² + 66.35² = 83.15² at arctan(-50.11/66.35) = -37° = 323°

    Impact speed = 83m/s

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