A geyser sends a blast of water in the air with a velocity of 96 feet per second. Find the height of the water after 4 seconds. What is the max height of the water? How long is the water in the air?

Answer 1

h(t) = -16t^2 + 96t

Plug in t = 4 and evaluate.

Maximum height is at the vertex of the height-vs.-time parabola, which is at

t = -96/[2(-16)] sec = ? sec

Evaluate h(t) at this value of t to get maximum height.

Since the water takes off from the ground, the total time in the air is just twice the time to maximum height.