Physics
posted by Addie on .
A stone is thrown straight up with an initial speed of 80 ft/s. How high does the stone go, and how long does it stay in the air?

It reaches a height H such that
M g H = (1/2) M Vo^2, therefore
H = Vo^2/(2g)
g = 32.2 ft/s^2 Vo is the initial velocity
H = 99.4 feet
Time going up = Vo/g = 2.48 s
The time coming down is the same.
Total time in the air = 4.96 s