If a projectile is fired with an initial velocity of v0 meters per second at an angle α above the horizontal and air resistance is assumed t

Question

If a projectile is fired with an initial velocity of v0 meters per second at an angle α above the horizontal and air resistance is assumed to be negligible, then its position after t seconds is given by the parametric equations x = (v0 cos(α))t y = (v0 sin(α))t − 1 2 gt2 where g is the acceleration due to gravity (9.8 m/s2). (Round your answers to the nearest whole number.) (a) If a gun is fired with α = 30° and v0 = 500 m/s. When will the bullet hit the ground?

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Thông Đạt 3 years 2021-08-22T04:45:42+00:00 1 Answers 142 views 0

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    2021-08-22T04:46:42+00:00

    Answer:

    a) The bullet hits the ground 51.02 s after it was fired.

    b) 22,092.3 units

    c) 3188.8 units

    Explanation:

    a) Assuming that the level the bullet was fired from is the ground level.

    The bullet hits the ground when y = 0

    y = (v₀ sin(α))t − (1/2) gt²

    v₀ = 500 m/s

    α = 30°

    y = 0

    0 = (500 sin 30) t – 0.5(9.8)t²

    4.9t² – 250t = 0

    t(4.9t – 250) = 0

    t = 0 s or (4.9t – 250) = 0

    The t = 0 s indicates that the bullet was indeed fired from the ground level.

    The time it eventually hits the ground back

    4.9t = 250

    t = 51.02 s

    The bullet hits the ground 51.02 s after it was fired.

    b) The distance from the firing point that the bullet lands.

    x = (v₀ cos(α))t

    At this horizontal distance, t = 51.02 s

    Substituting the parameters

    x = (500 cos 30°) × 51.02

    x = 22,092.3 units

    c) Maximum height attained by the bullet

    Maximum height is given by

    H = (u² sin² α)/2g

    H = (500² sin² 30°)/(2×9.8)

    H = 3188.8 units

    Hope this Helps!!!

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