A projectile is thrown upward so that its distance is above the ground after t seconds is given by the function h(t)=-16t^2+576t. After how many seconds does the projectile reach its maximum height?

Question

anonymous · Answer

Kirby can you help  me?
@kirbykirby

kirbykirby · Answer

oh yes sorry I got distracted by another question.. I'm almost done typing

kirbykirby · Answer

If you don't know calculus, then you can do this knowing that this is a quadratic function, so the maximum height occurs at the vertex (this is because you have a parabola of the forma $at^2+bt+c$, where $a<0$ (here c = 0), so the parabola looks like this $\cap$. 

So an easy way to handle this is to manipulate your equation into vertex form:
$-16t^2+576 t \ =-16(t^2-36t)$ 
complete the square by adding and subtracting $\left(\frac{b}{2}\right)^2=\left(\frac{-36}{2}\right)^2=324$

$=-16(t^2-36t+324-324)\
=-16(t-18)^2-16(-324)\
=-16(t-18)^2+5184$

Your parabola is of the form $a(x-h)^2+k$  now, so the hack occurs at the vertex. So the maximum height is 5184 at t=18

kirbykirby · Answer

so the max occurs at the vertex*

anonymous · Answer

thank you so much, I actually found a older post of the same question where ganeshie8 answered exactly the same! Yours is clearer though and I really appreciate the help!

kirbykirby · Answer

oh awesome :) ! your welcome =]