Answer:
1. ![h(t)=-16t^2+15t+6.5](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B15t%2B6.5)
<em>2. The ball hits the ground at t = 1.26 seconds</em>
<em>3. The basketball reaches its maximum height at t=0.47 seconds</em>
<em>4. The maximum height of the ball is 3.52 feet</em>
Step-by-step explanation:
<u>Function Modeling</u>
Reality can sometimes be modeled by mathematics. Functions are a great tool to explain the behavior of the measured magnitudes, it can also be used to predict future values and help to make decisions.
We are given a function to model the height (in feet) of a basketball once it's shot from the player. The function is
![h(t)=-16t^2+v_ot+h_o](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2Bv_ot%2Bh_o)
where t is the time in seconds,
the initial speed and
the initial height. We are also given the values
Part 1
![v_o=15\ ft/s,\ h_o=6.5\ ft](https://tex.z-dn.net/?f=v_o%3D15%5C%20ft%2Fs%2C%5C%20h_o%3D6.5%5C%20ft)
The complete model is
![h(t)=-16t^2+15t+6.5](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B15t%2B6.5)
Part 2
To find the time the basketball hits the ground, we must set its height to zero:
![-16t^2+15t+6.5=0](https://tex.z-dn.net/?f=-16t%5E2%2B15t%2B6.5%3D0)
To solve this quadratic equation, we'll use the solver formula
![\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
where a = -16, b = 15, c = 6.5
![\displaystyle t=\frac{-15\pm \sqrt{15^2-4\times (-16)\times 6.5}}{2(-16)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7B-15%5Cpm%20%5Csqrt%7B15%5E2-4%5Ctimes%20%28-16%29%5Ctimes%206.5%7D%7D%7B2%28-16%29%7D)
![\displaystyle t=\frac{-15\pm 25.32}{-32}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7B-15%5Cpm%2025.32%7D%7B-32%7D)
This produces two solutions:
![t=-0.322\ sec,\ t=1.26\ sec](https://tex.z-dn.net/?f=t%3D-0.322%5C%20sec%2C%5C%20t%3D1.26%5C%20sec)
We discard the negative solution because time cannot be negative, thus the ball hits the ground at t = 1.26 seconds
Part 3
A quadratic function of the form
![at^2+bt+c](https://tex.z-dn.net/?f=at%5E2%2Bbt%2Bc)
has its extrema value (maximum or minimum) at
![\displaystyle t=-\frac{b}{2a}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D-%5Cfrac%7Bb%7D%7B2a%7D)
If a>0, it's a maximum, otherwise it's a minimum
. Since a=-16, we'll get a maximum.
Computing the value of t to make the height be maximum
![\displaystyle t=-\frac{15}{2(-16)}=0.47\ sec](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D-%5Cfrac%7B15%7D%7B2%28-16%29%7D%3D0.47%5C%20sec)
The basketball reaches its maximum height at t=0.47 seconds
Part 4
The maximum height can be computed by using the function of h evaluated in t=0.47 sec
![h(t)=-16t^2+15t+6.5](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B15t%2B6.5)
![h_m=-16(0.47)^2+15\times 0.47+6.5](https://tex.z-dn.net/?f=h_m%3D-16%280.47%29%5E2%2B15%5Ctimes%200.47%2B6.5)
![h_m=3.52\ feet](https://tex.z-dn.net/?f=h_m%3D3.52%5C%20feet)
The maximum height of the ball is 3.52 feet