A rocket 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.

y = 17x - 4.5x²

height = 16m at 2 sec

total time = 3.77 sec

Step-by-step explanation:

Given the initial velocity of rocket = 56ft/sec

We use newton formula

s = ut + 1/2at²where s = distance

u = initial velocity

t = time

a = acceleration due to gravity = 9.8m/s²

since the rocket is launched upward, it is against the gravity so a = - 9.81.

56 feet = 17.0688 = 17

suppose s = height = y and time = x

y = 17x - 1/2 (9.81) x²

or

y = 17x - 4.5x²

b)

as x can not be negative so domain will be from 0 to 3.78

put value of x to find value of y and then plot these on graph

This graphs will be a parabola which opens downward.

c)

at 2 sec, y value is greatest = 16 meters

time for which rocket was in the air = 3.77sec