Question

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?

Answer 1

Answer:

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:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

Height of the diver after t seconds:

$h(t) = -4.9t^2 + 12t + 5$

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

Quadratic equation with $a < 0$, so the parabola is concave down, and it will be above 15m between the two roots that we found for $h(t) = 15$. So

$h(t) = -4.9t^2 + 12t + 5$

$15 = -4.9t^2 + 12t + 5$

$-4.9t^2 + 12t - 10 = 0$

Quadratic equation with $a = -4.9, b = 12, c = -10$. Then

$\Delta = 12^{2} - 4(-4.9)(-10) = -52$

Negative $\Delta$, 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 $h(t) = 0$. So

$h(t) = -4.9t^2 + 12t + 5$

$0 = -4.9t^2 + 12t + 5$

$-4.9t^2 + 12t + 5 = 0$

Quadratic equation with $a = -4.9, b = 12, c = 5$. Then

$\Delta = 12^{2} - 4(-4.9)(5) = 242$

$x_{1} = \frac{-12 + \sqrt{242}}{2*(-4.9)} = -0.36$

$x_{2} = \frac{-12 - \sqrt{242}}{2*(4.9)} = 2.81$

Time is a positive measure, so we take 2.81.

The stunt diver is in the air for 2.81 seconds.