Question

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 can be modeled by h=-16t² + 70t. How long is the boiling water in the air?

Answer 1

To find how long the boiling water of a geyser is in the air, we solve the equation h=-16t² + 70t for t when h=0. Factoring the equation gives t=0 and t=70/16, indicating the water is in the air for about 4.375 seconds.

Step-by-step explanation:

The student's question involves determining how long boiling water from a geyser is in the air, using the quadratic equation h=-16t² + 70t to model the height over time. The water is in the air from the time it is shot upwards at t=0 until it comes back down to the ground when h=0.

To find out how long the water is in the air, we need to solve the equation for t when h=0. This implies:
0 = -16t² + 70t

This can be factored to give:
t(70 - 16t) = 0

Setting each factor equal to zero provides two solutions, t=0 (when the water is shot upwards) and t=70/16 (when the water lands back on the ground). The latter is the answer to the question, which in decimal form is approximately 4.375 seconds. Therefore, the water is in the air for about 4.375 seconds.