Projectile is thrown upward so that its distance above the ground after t seconds is given by the function h(t) = - 16t2 + 704t

maximum-height

#1

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:

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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)