The path of the ball can be represented by h=-16t^2+27t+3

Question

What is the maximum height of the ball?

How long will it take the ball to reach the ground?

Answer 1

h' = -32t +27 = 0

t = 27/32

max h = -16(27/32)^2 + 27(27/32) +3 = -(27^2)/64+ 27^2/32 +3 =

(27^2)/64 +3

=14.39 feet high

but on desmos graph 14.39 is the max height.

try rewriting the quadratic in vertex form and find the y coordinate of the vertex

-16(x^2 -27/16 + (27/16)^2) + (27^2)/64+3

-16t^2 +27t +3 = 0

t = 1.79212 seconds to hit the ground