A person standing close to the edge on top of a 144-foot building throws a ball vertically upward. The quadratic function h ( t ) = − 16 t 2 + 128 t + 144 models the ball's height about the ground, h ( t ) , in feet, t seconds after it was thrown. What is the maximum height of the ball?

Step-by-step explanation:

Given the quadratic equation that models the height of the ball above the ground as;

h ( t ) = − 16 t^2 + 128 t + 144

t is in seconds

To get the maximum height of the ball, first we must know the time taken by the ball to reach the maximum height.

Note that the ball has a velocity of zero at its maximum height.

v(t) = dh/dt

v(t) = -32t+ 128

at max height, v(t) = 0

0 = -32t + 128

32t = 128

t = 128/32

t = 4secs

To get the maximum height of the ball, you will substitute t = 4 into the modeled function as shown;

h ( t ) = − 16 t^2 + 128 t + 144

h ( 4 ) = − 16 (4)^2 + 128 (4) + 144

h ( 4 ) = − 256 + 512 + 144

h(4) = 400m

Hence the maximum height of the ball is 400m

show us the graphs dude

step-by-step explanation: