A trained stunt diver is diving off a platform that is 15 m high into a pool of water that is 45 cm deep. The height, h, in meters, of the stunt diver above the water, is modeled by h=-4.9t^2+12t+5, where t is the time in seconds after starting the dive. a) How long is the stunt diver above 15 m?

b) How long is the stunt diver in the air?

a) 0 seconds.

b) The stunt diver is in the air for 2.81 seconds.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

.

This polynomial has roots such that , given by the following formulas:

Height of the diver after t seconds:

a) How long is the stunt diver above 15 m?

Quadratic equation with , so the parabola is concave down, and it will be above 15m between the two roots that we found for . So

Quadratic equation with . Then

Negative , which means that the stunt diver is never above 15m, so 0 seconds.

b) How long is the stunt diver in the air?

We have to find how long it takes for the diver to hit the ground, that is, t for which . So

Quadratic equation with . Then

Time is a positive measure, so we take 2.81.

The stunt diver is in the air for 2.81 seconds.

hope this !

its   c, f, a

step-by-step explanation: