Question 224730
what is the initial velocity of a geyser's boiling water this is the equation
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h=-16t+70t
I think it's h(t) = -16t^2 + 70t
h'(t) = -32t + 70
h'(0) = 70
The initial velocity = 70 ft/sec
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what is the maximum height of the boiling water?
The time of max height is half of the flight time.  From below, total time = 4.375 seconds.
1/2 is 2.1875 seconds the time at apogee
h(2.1875) = -16*2.1875^2 + 70*2.1875
max h = 76.5625 feet
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how long was the boiling water in the air?
Find when it's at h = 0. There will be 2 times, when it left and when it returns.
-16t^2 + 70t = 0
t = 0 (eruption)
t = 70/16 seconds
t = 4.375 seconds
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It's assumed that there's no evaporation and no air resistance.