Answer:
A) The initial height is 32
B) The higher height that the ball gets is 68
C) at 
Step-by-step explanation:
A) For this part you only have to evaluate the function in t=0, this is:

B) To obtain the higher height you have to find the maximum value that reaches the function
. For this we can use the first derivative rule.
Then we have:

How we only have one extreme point we assume that this is a maximum, and this point is in the value
, this is
. Then, evaluate the function
in 
we have

C) In this case we need to know when the function is 0. For this we can use the quadratic formula, with ![a=-16,\ b=48,\ c=32[\tex] and taking the positive solution.[tex]x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-(48) \pm \sqrt{b(48)^2-4(-16)(32)}}{2(-16)} = \frac{-48\pm \sqrt{2304+2048}}{-32}\\\\=\frac{-48+\sqrt{4352}}{-32}\approx 3.56](https://tex.z-dn.net/?f=a%3D-16%2C%5C%20b%3D48%2C%5C%20c%3D32%5B%5Ctex%5D%20and%20taking%20the%20positive%20solution.%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%5Btex%5Dx_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%3D%20%5Cfrac%7B-%2848%29%20%5Cpm%20%5Csqrt%7Bb%2848%29%5E2-4%28-16%29%2832%29%7D%7D%7B2%28-16%29%7D%20%3D%20%5Cfrac%7B-48%5Cpm%20%5Csqrt%7B2304%2B2048%7D%7D%7B-32%7D%5C%5C%5C%5C%3D%5Cfrac%7B-48%2B%5Csqrt%7B4352%7D%7D%7B-32%7D%5Capprox%203.56)