Question 724396: A ball is thrown upward with an initial velocity of 343 m/sec. When will it be 4900 meters high? I have been trying to help my son with his Algebra. I am having such a hard time. Could some one please show and explain to me an eqation to figure this problem out?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A ball is thrown upward with an initial velocity of 343 m/sec. When will it be 4900 meters high?
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The equation for ht vs. time should be given with the problem.
If not, the commonly used is
h(t) = -4.9t^2 + vt + C, t in seconds, h & C in meters
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h(t) = -4.9t^2 + 343t = 4900
4.9t^2 - 343t + 4900 = 0
t^2 - 70t + 1000 = 0
(t - 20)*(t - 50) = 0
t = 20 seconds going up
t = 50 seconds falling
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That's a good throw, ~ 767 mi/hr
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