**projectile is thrown upward so that its distance above the ground after t seconds is given by the function h(t) = - 16t2 + 704t. After how many seconds does the projectile take to reach its maximum height?**

**Answer:**

if you factor out h(t), it gives you: h(t)=−16t∗(t−44).

To get the maximum height, you have to solve for the time halfway between when it was launched (t = 0) and when it hits the same height as when it was launched or the “ground level” (t = 44). This means your answer is t = 22 seconds. (sorry for the confusion but this should be right)