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A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 14.5 m/s, and the distance from the limb to the level of the saddle is 3.02 m.

(a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? in meters

(b) For what time interval is he in the air?

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Added Wed, 11 Nov '15
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As the ranch hand drops vertically, his vertical velocity increases 9.8 m/s each second.

Distance = 1/2 * acceleration * time^2

3.02 = 1/2 * 9.8 * t^2

t = 0.785 seconds

(b) TIME IN THE AIR= 0.785 seconds

During the 0.785 seconds, the horse is running at a constant speed of 14.5 m/s toward the tree limb.

Distance = velocity * time

Distance = 14.5 m/s * 0.785 s = 11.3825 meters

(a) The horizontal distance between the saddle and limb when the ranch hand makes his move = 11.3825

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Added Wed, 11 Nov '15
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