I need help with this equation in the Description

Question

You are standing in a park and kick a ball straight up into the air. The ball travels at an initial velocity of 64 feet per second. The equation h(t)=-16t^2+64t models the height of the ball "h" at any given time "t". How long will the ball be in the air? When will the ball be at its maximum height? How high does the ball get?

Answer 1

h(t) = -16t² + 64t

x value of the vertex of parabola is at -b/2a = -64/(2(-16)) = -64/-32 = 2 sec.

Maximum is at -16(4) +64(2) = -64 + 128 = 64 ft.

Factoring to find the zeros: divide by -16: h = t² - 4t = 0 → t(t - 4) = 0

so t = 0 and t = 4 sec.

Suggest you graph the function on Desmos Graphing Calculator website.

Answer 2

h = -16t^2 +64t

take the derivative and set it equal to zero

v= h' = -32t +64 = 0

t =2 seconds to reach max height when velocity = 0

max height = -16(2^2) +64(2) = 64 feet high

the ball will be in the air 2(2) = 4 seconds, when it hits the ground