Problem: An object thrown upward slows to 2 meters/second, and subject only to the force of gravity, at a certain instant. .5000 seconds later, how fast will it be going and what will be its net displacement since the instant at which it attained the 2 m/s velocity?
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Solution: The object will accelerate downward at 9.8 m/s/s for .5000 sec, resulting in change of velocity (9.8 m/s/s)( .5000 sec) = 4.9 m/s in the downward direction. The final downward velocity will thus be 4.9 m/s - 2 m / s = 6.9 m/s (note that the upward initial velocity decreases the final downward velocity so must be subtracted). The average velocity during the .5000 seconds will thus be (- 2 m/s + 6.9 m/s)/2 = 4.45 m/s, and the net displacement will be ( .5000 sec)( 4.45 m/s) = 2.225 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.
Note that this analysis doesn't worry about whether v0y is upward or downward. Once the positive direction is declared, the sign of each quantity will reveal whether it is positive or negative.
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