Problem: An object moves both horizontally and vertically. Its horizontal motion is at a constant velocity of 99 m/s; its vertical motion begins at upward velocity 1.456 m/s and accelerates downward at a constant rate of 9.8 m/s/s. How far does it travel horizontally, and what is its downward vertical displacement, after 2.08 seconds?
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Solution: In 2.08 seconds, moving at 99 m/s, the object will move ( 99 m/s)( 2.08 sec) = 205.9 meters horizontally. Its vertical velocity will change by (9.8 m/s/s)( 2.08 sec) = 20.38 m/s in the downward direction. The final velocity will thus be 18.92 m/s, and its average downward velocity will be (- 1.456 m / s + 18.92 m/s)/2 = 8.732 m/s. The distance traveled will therefore be ( 8.732 m/s)( 2.08 sec) = 18.16 meters.
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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