Answer:
a) t = 2 seconds
b) 70 feet
Explanation:
First of all, we are going to take the derivative of the function (which is a displacement function) to make find its velocity function.
![h(t) = -16t^(2) +64t+6](https://img.qammunity.org/2022/formulas/mathematics/high-school/cxd85fzp635th8gapn755y9rw98v3z6mmm.png)
To find the derivative of the equation, we must multiply the coefficient by the exponent and then minus one from the exponent for each term.
![v(t)=(-16)(2)t^(2-1) +64(1)t^(1-1)+6(0)t^(0-1) \\v(t) = -32t +64](https://img.qammunity.org/2022/formulas/mathematics/high-school/sgkqgs730b0g1gyz1k4ikso4ziy2nxkeqb.png)
Now we can substitute in v=0
This is because the ball is at the top of its parabola path, it is the turning point and the slope is 0. The velocity physically speaking would also be 0 at the very top of its path.
![0=-32t+64](https://img.qammunity.org/2022/formulas/mathematics/high-school/9ffr5ebp09055a6s7i5nt7hw84u1qfue91.png)
Rearrange and we get:
![32t = 64\\t=2](https://img.qammunity.org/2022/formulas/mathematics/high-school/37d49tgqcfna4ppwdxcs5uvmvxziydkbo9.png)
Ball attains maximum height at 2 seconds.
To find the maximum height, we now just have to substitute back in the time into the original equation to find the height.
![h(t) = -16t^(2) +64t+6](https://img.qammunity.org/2022/formulas/mathematics/high-school/cxd85fzp635th8gapn755y9rw98v3z6mmm.png)
![h(2) = -16(2)^(2) +64(2)+6\\h(2)=-64+128+6\\h(2) = 70](https://img.qammunity.org/2022/formulas/mathematics/high-school/n36hwsbhg7qfh7klfdab8gy9pm7jy9cegf.png)
The maximum height attained is 70 feet.
Hope this helped!