Answer:
a) The initial height of the rock is of 48 feet.
b) It takes 1 seconds for the rock to reach maximum height.
c) The rock's maximum height is 50 feet.
d) It takes 6 seconds for the rock to land on the ground.
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
![f(x) = ax^{2} + bx + c](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ax%5E%7B2%7D%20%2B%20bx%20%2B%20c)
It's vertex is the point ![(x_{v}, f(x_{v})](https://tex.z-dn.net/?f=%28x_%7Bv%7D%2C%20f%28x_%7Bv%7D%29)
In which
![x_{v} = -\frac{b}{2a}](https://tex.z-dn.net/?f=x_%7Bv%7D%20%3D%20-%5Cfrac%7Bb%7D%7B2a%7D)
If a<0, the vertex is a maximum point, that is, the maximum value happens at
, and it's value is ![f(x_{v})](https://tex.z-dn.net/?f=f%28x_%7Bv%7D%29)
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:
![x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D%20%5Cfrac%7B-b%20%2B%20%5Csqrt%7B%5Cbigtriangleup%7D%7D%7B2%2Aa%7D)
![x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%20%5Cfrac%7B-b%20-%20%5Csqrt%7B%5Cbigtriangleup%7D%7D%7B2%2Aa%7D)
![\bigtriangleup = b^{2} - 4ac](https://tex.z-dn.net/?f=%5Cbigtriangleup%20%3D%20b%5E%7B2%7D%20-%204ac)
In this question:
The height of the ball after t seconds, in feet, is given by:
![h(t) = -2t^2 + 4t + 48](https://tex.z-dn.net/?f=h%28t%29%20%3D%20-2t%5E2%20%2B%204t%20%2B%2048)
Which is a quadratic equation with ![a = -2, b = 4, c = 48](https://tex.z-dn.net/?f=a%20%3D%20-2%2C%20b%20%3D%204%2C%20c%20%3D%2048)
a) What is the initial height?
This is
. So the initial height of the rock is of 48 feet.
b) How many seconds does it take to reach the maximum height?
This is the value of t at the vertex. So
![t_{v} = -\frac{b}{2a} = -\frac{4}{2(-2)} = 1](https://tex.z-dn.net/?f=t_%7Bv%7D%20%3D%20-%5Cfrac%7Bb%7D%7B2a%7D%20%3D%20-%5Cfrac%7B4%7D%7B2%28-2%29%7D%20%3D%201)
It takes 1 seconds for the rock to reach maximum height.
c) What is the rock's maximum height?
This is the height of the ball after 1 second. So
![h(1) = -2(1)^2 + 4(1) + 48 = 50](https://tex.z-dn.net/?f=h%281%29%20%3D%20-2%281%29%5E2%20%2B%204%281%29%20%2B%2048%20%3D%2050)
The rock's maximum height is 50 feet.
d) How many seconds does it take to for the rock to land on the ground?
This is t for which
, so we solve the quadratic equation.
![\bigtriangleup = (4)^2-4(-2)(48) = 400](https://tex.z-dn.net/?f=%5Cbigtriangleup%20%3D%20%284%29%5E2-4%28-2%29%2848%29%20%3D%20400)
![t_{1} = \frac{-4 + \sqrt{400}}{2*(-2)} = -4](https://tex.z-dn.net/?f=t_%7B1%7D%20%3D%20%5Cfrac%7B-4%20%2B%20%5Csqrt%7B400%7D%7D%7B2%2A%28-2%29%7D%20%3D%20-4)
![t_{2} = \frac{-4 - \sqrt{400}}{2*(-2)} = 6](https://tex.z-dn.net/?f=t_%7B2%7D%20%3D%20%5Cfrac%7B-4%20-%20%5Csqrt%7B400%7D%7D%7B2%2A%28-2%29%7D%20%3D%206)
Time is a positive measures, so it takes 6 seconds for the rock to land on the ground.