Posted by liz on Friday, March 30, 2012 at 12:09am.
a) If the time in the air is t, it spends t/2 going up and t/2 coming back down. The distance it falls in time t/2 is
(1/2)*g*(t/2)^2 = gt^2/8
b) gt^2/8 = 9.8* 16/8 = 19.6 m
c) The formula linking max height to time in the air is independent of the angle thrown. So the answer is "yes".
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