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1 September, 10:12

An archer shoots an arrow at a 75.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

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  1. 1 September, 11:30
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    a) Θ = 18.5°

    b) h = 6.26 m, 3.50 m, arrow goes over the branch

    Explanation:

    having the following dа ta:

    Vo = 35 m/s

    Θ = ?

    horizontal distance = 75 m

    a)

    using the following equations:

    Voy = Vo * (sin Θ) = 35 * (sin Θ)

    Vox = Vo * (cos Θ) = 35 * (cos Θ)

    horizontal distance to target = 75 = Vox * (2t); where t = Voy/g

    replacing values:

    75 = Vox * (2/g) * (Voy) = 2 * (Vox) * (Voy) / 9.8 = (Vo) ²*[2 * (sin Θ) * (cos Θ) ]/g = (Vo) ² (sin2Θ) / g

    solving and using trigonometric identities:

    sin2Θ = 75 * (g) / (Vo) ² = 75 * (9.8) / (35) ² = 0.6

    2Θ = 36.91°

    Θ = 18.5°

    b)

    The time to reach the maximum height will be equal to:

    t = Voy/g = 35 * (sin18.5°) / 9.8 = 1.13 s

    and the maximum height will be equal to:

    h = 1/2gt² = (0.5) * (9.8) * (1.13) ² = 6.26 m, 3.50 m, arrow goes over the branch
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