Problem: An object moves both horizontally and vertically. Its horizontal motion is at a constant velocity of 21.95 m/s; its vertical motion begins at downward velocity 19.8 m/s and accelerates downward at a constant rate of 9.8 m/s/s, until the object strikes the ground 19.75 meters below its starting point. How far does the object travel horizontally?
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Solution: We find the time of fall by solving the equation `ds = v0 * `dt + .5 a `dt ^ 2 for `dt. We find that the acceptable solution is `dt = .8278 sec. Moving horizontally at 21.95 m/s for this time the distance covered will be 18.17 m.
Since acceleration is uniform, the average velocity will therefore be
ave y velocity = vyAve = (v0y + vfy) / 2 = (v0y + (v0y + g `dt)) / 2 = (2 v0y + g `dt) / 2.
The vertical displacement will therefore be
vertical displacement = `dsy = vyAve * `dt = (2 v0y + g `dt) / 2 * `dt.
The horizontal displacement at constant horizontal velocity v0x will be dsx = v0x * `dt.
Recall that the horizontal and vertical motions are independent except for the common time interval, with horizontal velocity being constant (i.e., horizontal acceleration is 0) and vertical acceleration being constant (i.e., vertical acceleration is the acceleration of gravity).
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