Question

Damon thank you for the previous help.

A geyser sends a blast of boiling water high into the air. During the eruption, the height (h) (in feet) of the water (t)seconds after being forced out from the ground could be modeled by h = -16t*t + 70t.

The last two questions are: how do you find the maximum height of the boiling water and how long is the boiling water in the air ? (using the quadratic formula)

Answer 1

h = -16 t^2 + 70 t

The maximum height can be found by calculus, by graphing h vs. t or "completing the square". They probably want you to complete the square. That means changing the equation to:
h = -16t^2 + 70t - 35 + 35
= -16(t^2 -70t/16 + 35^2/16^2)+ 35^2/16
= -16(t - 35/16)^2 + 35^2/16
The highest value of h occurs when t = 35/16, in which case the squared term becomes zero. The maximum height then is 35^2/16 = 76.6 feet

The water is in the air until h = 0 a second time. If you use the quadratic formula (with a = -16, b = 70 and c = 0), the answer is written

t = [-70 - sqrt(70^2]/-32 = 140/32 = 4.375 s