Question

A projectile is thrown upward so that it's distance above the ground after t seconds is h=-16t^2 +672t. After how many seconds does it reach its maximum height?

Answer 1

Answer: 21 seconds

Step-by-step explanation:

We know that a parabola with equation $y=ax^2+bx+c$ attains its maximum height at :-

$x=\dfrac{-b}{2a}$

The given function: $h=-16t^2 +672t$

It will attain its maximum height at :

$t=\dfrac{-672}{2(-16)}=21$

Hence , after 21 seconds the projectile will reach its maximum height .

Answer 2

It will reach its maximum height at the vertex of the parabola.  Which is always:

(-b/(2a), (4ac-b^2)/(4a))

So the time when this occurs is the x coordinate, or t in this case.

-b/(2a), and b=672 and a=-16 so

t=-672/-32

t=21 seconds