Physics
posted by Tomtom .
You stick your arm over the edge of the Grand Canyon, and throw a ball vertically upwards with a velocity of 15 m/s. Compared to the heighht that you throw the ball, what height is the ball after 6 seconds?

t(up) = (Vf  Vo) / g,
t(up) = (0  15) / 9.8 = 1.53s. to reach max. ht.
hmax = (Vf^2  Vo^2) / 2g,
hmax = (0  (15)^2) / 19.6 = 11.5m.
above launching point.
h = Vo*t + 0.5g*t^2,
h = 15*6  4.9*6^2 = 86.4m = The distance below max. ht.
Freefall distance=86.4  11.5 = 98m
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