Question

a ball is thrown into the air with an upward velocity of 36 ft/s. it’s height h in feet after t seconds is given by the function h=-16t^2+36t+9. in how many seconds does the ball reach its maximum height?

Answer 1

Complete the square:

$h(t)=-16t^2+36t+9=-16\left(t^2-\dfrac94t\right)+9=-16\left(\left(t-\dfrac98\right)^2-\dfrac{81}{64}\right)+9$

$h(t)=-16\left(t-\dfrac98\right)^2+\dfrac{117}4$

That is, the graph of $h(t)$ is a parabola with vertex at $(t,h(t))=\left(\dfrac98,\dfrac{117}4\right)$, meaning the ball reaches its maximum height of $\dfrac{117}4=29.25\,\mathrm{ft}$ after $t=\dfrac98=1.125\,\mathrm s$ of being thrown.