1. Velocity and maximum height

A ball is thrown upward from a height of 32 ft with an initial velocity of 16 ft/sec. Find a function that gives the position of the ball at time t. What is the maximum height the ball will reach? [Recall: a(t)= -32 ft/sec squared].

I need some help answering this for a test I took. We get points back on our exams if we correct our mistakes. Thanks guys.

2. height(t) = 0.5a^2 + vt + initial height (standard height equation)

a = acceleration (-32 ft/sec^2)

v = initial velocity

Differentiate, find critical number, then use either the first or second derivative test to find the maximum height.

Edit: don't bother using the derivative tests because there's only one critical number lol.

3. So the function that gives the position at time t would be

0.5(-32 ft/sec^2) + (16 ft/sec)(t) + 32 ft ?

4. $\int a(t)dt = v(t)$
$\int -32dt = -32t + C$

C = Initial velocity = 16

$\int v(t)dt = s(t)$
$\int (-32t + 16)dt = -16t^2 + 16t + C$

C = initial height = 32

$s(t) = -16t^2 + 16t + 32$

Maximum height can be found by calculating t when velocity changes from positive to negative.