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 10.0 m/s, and the distance from the limb to the level of the saddle is 3.00 m. (a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? (b) For what time interval is he in the air?

Question

Physics

Brice Howard

23 July, 12:25

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 10.0 m/s, and the distance from the limb to the level of the saddle is 3.00 m. (a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? (b) For what time interval is he in the air?

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Landon Walter · Answer 1

The time it will take him to fall can be found from:-3m = - (g*t^2) / 2

Find that time, it's the time the horse will travel horizontal while the cowboy is falling. So the horizontal distance away is 10 m/s * t