Aball is thrown vertically upward from the ground with an initial velocity of 122 ft/sec. use the quadratic function
h(t) = -16t2 + 122t +0 to find how long it will take for the ball to reach its maximum height, and then find the
maximum height. round your answers to the nearest tenth.

232.6 metres after 3.8 seconds.

Step-by-step explanation:

h(t) = -16t² + 122t

a = -16  b = 122  c = 0

Substitute into the quadratic formula

(Ignore the Â)

Split the equation at the ±

The two x-intercepts at 0 and 61/8. The midpoint of the x-intercepts is the axis of symmetry, which is the x-coordinate of the vertex.

Midpoint = [0 + (61/8)] / 2

Midpoint = 61/16   <= This is the time for maximum height

t = 61/16

t = 3.8125  => Round to t = 3.8

To find the maximum height, substitute t=61/16 into the equation

h(t) = -16t² + 122t

h(61/16) = -16(61/16)² + 122(61/16)

h(61/16) = 232.5625  => Round to h = 232.6

Therefore, the ball will reach the maximum height of 232.6 metres after 3.8 seconds.

answer: 10 meter

step-by-step explanation:

need details

what statement? is there an image?