Problem: An object initially moving downward at 6 m/s falls freely near the surface of the earth. How fast will it be moving .3000 seconds later, and what will be its downward displacement during that time?
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Solution: The object will accelerate downward at 9.8 m/s/s for .3000 sec, resulting in change of velocity (9.8 m/s/s)( .3000 sec) = 2.94 m/s in the downward direction. The final velocity will then be 6 m / s + 2.94 m/s = 8.940 m/s. The average downward velocity during the .3000 seconds will be ( 6 m/s + 8.940 m/s)/2 = 7.47 m/s, and the net displacement will be ( .3000 sec)( 7.47 m/s) = 2.241 m.
Generalized Response: The object has initial vertical velocity v0y and acceleration ay = g. In time interval `dt the velocity will change by `dv = g `dt, resulting in velocity vfy = v0y + g `dt.
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. It doesn't matter. As long as a positive direction is chosen (the positive direction could be either up or down) and every quantity given the appropriate sign (e.g., if the positive direction is down then the acceleration is positive and upward velocities are negative, with positive displacements being downward; while if the positive direction is down then the acceleration is negative and upward velocities are positive, with negative displacements being downward), the sign of each quantity will determine whether it is upward or downward.
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