# An object is thrown upward from the ground with an initial velocity of 32ft/s. What is the maximum height the object obtains using the formula s = -16t^2 + 32t, where s = distance above the ground in feet, and t= time in seconds?

Apr 14, 2015

The maximum height with respect to time will occur when the derivative of the distance(time) function equals $0$

$s = - 16 {t}^{2} + 32 t$

$\frac{\mathrm{ds}}{\mathrm{dt}} = - 32 t + 32$

Maximum occurs when
$- 32 t + 32 = 0$
$\rightarrow t = 1$

When $t = 1$ the object is at a height of
$- 16 {\left(1\right)}^{2} + 32 \left(1\right)$

$= 16$ (feet)