A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under a tree.
The constant speed of the horse is 10m/s, and the distance from the limb to the saddle is 3m. What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?
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This is a simple one. When the daring ranch hand decides to let go of the branch, he will start accelerating toward the Earth at a rate of 9.8 meters per second per second. That's almost as fast as the horse's velocity, which is a constant 10 meters per second. If the daring ranch hand has three meters to cover, distance-wise, to drop into the saddle, all he has to do is jump when the horse is at the seven meter mark. He and the horse will cover their repsective final three meters at the same velocity, 3, 2, 1, then they intersect! The daring young ranch hand lands squarely in the saddle, and both he and the horse continue on together. The force of gravity will cause him to travel downward just as fast as the horse is traveling underneath the branch. Their respective speeds are virtually a one to one ratio, so if there is three meters of distance to fall, when the horse has three meters to go to be directly underneath the branch, jump!
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