Question

A person standing close to the edge on the top of an 80-foot building throws a ball
vertically upward with an initial velocity of 64 feet per second. The function
s(t) = -16t 2 + 64t + 80 describes the ball's height above the ground, s(t), in feet, t
seconds after it is thrown.
(a) After how many seconds does the ball reach its maximum height?
(b) Find the maximum height of the ball.

Answer 1

Take the derivative, this will give you a velocity function.

v(t)= -32t+64

Solve for when the velocity is 0

0=-32t+64

Answer 2

how didu get these numbers

Answer 3

Derivative of each term of your position function

for example on derivatives:

derivative of x^2 = 2x
derivative of 3x^2= 6x
derivative of 4x=4
etc...