# A ball is thrown vertically with an initial velocity of 64 feet per second. Its height above the ground after t seconds is given by h(t) = 64t—16t^2. What is the maximum height?

Nov 17, 2015

I found ${h}_{\max} = 64 \text{feet}$

#### Explanation:

Ok...probably you can do this differently but I would try to find the vertex of the parabola describing the trajectory:
1) derive it:
h`(t)=64-32t
2) set derivative equal to zero:
$64 - 32 t = 0$
$t = \frac{64}{32} = 2 \sec$
3) use this value of $t$ into your trajectory:
$h \left(2\right) = {h}_{\max} 64 \cdot 2 - 16 \cdot 4 = 64 \text{feet}$

Graphically: