Problem: An object moves both horizontally and vertically. Its horizontal motion is at a constant velocity of 51.15 m/s; its vertical motion begins at upward velocity 20 m/s; the object accelerates downward at a constant rate of 9.8 m/s/s, until it strikes the ground and stopping 17.5 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 = 4.822 sec. Moving horizontally at 51.15 m/s for this time the distance covered will be 246.6 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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