Arocket is launched vertically from the ground with an initial velocity of 56 ft/sec

(a) write a quadratic function that shows the height, in feet, of the rocket t seconds after it was launched.
(b) graph on the coordinate plane.
(c) use your graph from part 3(b) to determine the rocket’s maximum height, the amount of time it took to reach its maximum height, and the amount of time it was in the air.

a) h(t) = -16t² + 56t; b) the graph is attached; c) maximum height is 49 feet, it takes 1.75 seconds to reach the maximum, and it is in the air for 3.5 seconds.

Step-by-step explanation:

Equations for these situations are in the form

h(t) = -16t² + v₀t + h₀, where t is the tie in seconds, v₀ is the initial velocity and h₀ is the initial height. The initial height in this case is 0, since it is launched from the ground; and the initial velocity is 56.  This gives us the function

h(t) = -16t² + 56t

The graph is attached.

Using the graph, we can trace the function to the maximum.  The y-value of the maximum, or the maximum height, is 49 feet.  The x-value of the maximum, the time it takes, is 1.75 seconds.

The roots of this equation are at x = 0 and x = 3.5.  This means the rocket is in the air for 3.5-0 = 3.5 seconds.

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.