Answer:
1.
![h(t)=-16t^2+15t+6.5](https://img.qammunity.org/2021/formulas/mathematics/high-school/3youfllx764q9h3wwljdg6pd4a3g1v5p12.png)
2. The ball hits the ground at t = 1.26 seconds
3. The basketball reaches its maximum height at t=0.47 seconds
4. The maximum height of the ball is 3.52 feet
Explanation:
Function Modeling
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://img.qammunity.org/2021/formulas/mathematics/high-school/n9jytxw5z4if52z0wapzxiyhuozz6sh7yg.png)
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://img.qammunity.org/2021/formulas/mathematics/high-school/4gvwn50rie1cdnz77az31f400ky5euowrd.png)
The complete model is
![h(t)=-16t^2+15t+6.5](https://img.qammunity.org/2021/formulas/mathematics/high-school/3youfllx764q9h3wwljdg6pd4a3g1v5p12.png)
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://img.qammunity.org/2021/formulas/mathematics/high-school/tdcy7zelwa7gt6wnglzradmui397ky7vit.png)
To solve this quadratic equation, we'll use the solver formula
![\displaystyle x=(-b\pm √(b^2-4ac))/(2a)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/s3age13u1k5t2be3r2834z0ijj7y78cct5.png)
where a = -16, b = 15, c = 6.5
![\displaystyle t=(-15\pm √(15^2-4* (-16)* 6.5))/(2(-16))](https://img.qammunity.org/2021/formulas/mathematics/high-school/gj4rrppiqu6up0pqpsl8iflant2ebdlgjc.png)
![\displaystyle t=(-15\pm 25.32)/(-32)](https://img.qammunity.org/2021/formulas/mathematics/high-school/lp9gp2hg5y511p91bdkk0m0oscbmt7y1tf.png)
This produces two solutions:
![t=-0.322\ sec,\ t=1.26\ sec](https://img.qammunity.org/2021/formulas/mathematics/high-school/z6nn9kxbyf6yfcewqolkirbhi1mbmbrs86.png)
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://img.qammunity.org/2021/formulas/mathematics/high-school/diuuldlmhrg16twiyaavm8nqswikjgw2bb.png)
has its extrema value (maximum or minimum) at
![\displaystyle t=-(b)/(2a)](https://img.qammunity.org/2021/formulas/mathematics/college/7omm6j9bc1erofer5h95w0p6rzkwpanyuu.png)
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=-(15)/(2(-16))=0.47\ sec](https://img.qammunity.org/2021/formulas/mathematics/high-school/uz8y087mpktqqkr3w7hz1l7phz248b59ru.png)
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://img.qammunity.org/2021/formulas/mathematics/high-school/3youfllx764q9h3wwljdg6pd4a3g1v5p12.png)
![h_m=-16(0.47)^2+15* 0.47+6.5](https://img.qammunity.org/2021/formulas/mathematics/high-school/dfeyoxtnjdfmokfkdob35c706m9wcfilfi.png)
![h_m=3.52\ feet](https://img.qammunity.org/2021/formulas/mathematics/high-school/5izgxc48ckt38t8dk33or3s1a813ge83fe.png)
The maximum height of the ball is 3.52 feet