A ball is shot out of a cannon. The height of the ball, h in meters, as a function of the time, t in seconds, since it was shot is given by the equation h= -5t^2 +30t +1. What is the maximum height?

Feb 23, 2016

$46 m$

Explanation:

The maximum and minimum values will be at points where the derivative of this function is zero, ie
$\exists {h}_{\min / \max} \iff \frac{\mathrm{dh}}{\mathrm{dt}} = 0$

$\iff - 10 t + 30 = 0$

$\iff t = 3 s$.

Since the second derivative $\frac{{d}^{2} h}{{\mathrm{dt}}^{2}} {|}_{t = 3} = - 10 < 0$,
it implies that $t = 3$ is a relative maximum of the function $h \left(t\right)$.

$\therefore {h}_{\max} = h \left(3\right)$

$= - 5 \left({3}^{2}\right) + 30 \left(3\right) + 1$

$= 46 m$.