an object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the object's maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t^2 + 64t +80?

Question

aaronq · Answer

The maximum height is when the velocity =0

The velocity function for the object is the first derivative w.r.t. time of the position function.

$v(t)=\dfrac{d[h(t)]}{dt}$

Find the time, t, when $v(t)=0$, use that time in $h(t)$ to find the height.