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18 April, 05:50

A missile is moving 1810 m/s at a 20.0 degree angle. It needs to hit a target 19,500 m away in a 32 degree direction in 9.20 seconds. What is the direction of the acceleration that the engine must produce?

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Answers (2)
  1. 18 April, 06:01
    0
    79.1°

    Explanation:

    A missile is moving 1810 m/s at a 20.0 degree angle. It needs to hit a target 19,500 m away in a 32 degree direction in 9.20 seconds. The direction of the acceleration that the engine must produce is 79.1°
  2. 18 April, 06:06
    0
    79.1°

    Explanation:

    Given:

    x₀ = 0 m

    y₀ = 0 m

    x = 19500 cos 32°

    y = 19500 sin 32°

    v₀x = 1810 cos 20°

    v₀y = 1810 sin 20°

    t = 9.20

    Find:

    ax, ay, θ

    First, in the x direction:

    x = x₀ + v₀ t + ½ at²

    19500 cos 32° = 0 + (1810 cos 20°) (9.20) + ½ ax (9.20) ²

    16537 = 15648 + 42.32 ax

    ax ≈ 21.01

    And in the y direction:

    y = y₀ + v₀ t + ½ at²

    19500 sin 32° = 0 + (1810 sin 20°) (9.20) + ½ ay (9.20) ²

    10333 = 5695 + 42.32 ay

    ay ≈ 109.6

    The direction of the acceleration is therefore:

    θ = atan (ay / ax)

    θ = atan (109.6 / 21.01)

    θ ≈ 79.1°
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