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1. A) An archer shoots an arrow at a 73.0 m distant target, the bull's-eye of which 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 37.0 m/s? (Although neglected here, the atmosphere provides significant lift to real arrows.)

__________�

(b) There is a large tree halfway between the archer and the target with an overhanging branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

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Added Thu, 28 Apr '16
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(a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 37.0 m/s?

Since the height of the bulls-eye is the same as release height of the arrow, you can use the range equation to determine the angle.

Range = v^2/g * sin 2?

73 = 37^2/9.8 * sin 2?

sin 2? = 73 * 9.8/37^2 = 715.5/1369

2? = Inverse sin of 715.5/1369

? = 0.5 * Inverse sin of 715.5/1369

The angle is approximately 15.75244?

(b) There is a large tree halfway between the archer and the target with an overhanging branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

We need to determine the height of the arrow when it is halfway between the archer and the target. At this horizontal distance, the arrow is at the middle of its path. This means the arrow is at its maximum height. As the arrow rises to its maximum height, its vertical velocity decreases from its initial vertical velocity to 0 m/s at the rate of 9.8 m/s each second. Use the following equation to determine the arrow�s maximum height.

vf^2 = vi^2 + 2 * a * d

vf = 0

vi = v * sin ? = 37 * sin 15.75244? ? 10.0448 m/s

a = -9.8 m/s^2

d = maximum height

0 = (37 * sin 15.75244?)^2 + 2 * -9.8 * d

-(37 * sin 15.75244?)^2 = -19.6 * d

d = (37 * sin 15.75244?)^2 � 19.6

The maximum height is approximately 5.1478 meters. Since this is higher than the overhanging branch, the arrow will go over the branch.

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Added Thu, 28 Apr '16
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