A rocket is launched vertically from the ground with an initial velocity of 64 ft/sec. Write a quadratic function that shows the height, in feet, of the rocket t seconds after it was launched. Graph on the coordinate plane.

We use the formula;  

s = ut + 1/2at²

s = distance

u = initial velocity

t= time

a = acceleration (in  case we use acceleration due to gravity = 10m/s²)

since the rocket is launched upward, it is decelerating, so the equation becomes;  

s = ut - 1/2at²

take s = height = y

t= time = x

64ft/sec = 19.5m/s

the equation becomes;  

y = 19.5x - 1/2(10)x²

or

y = -5x² + 19t

the graph is attached;  

from the graph

c) maximum height = 19m = 62.3ft

time taken to reach max. height = 1.95 sec

time the rocket was in air = 3.9 sec.

The correct graph is the second one, with the points being at (-6,0), (2,0) and reaching down to (-2,-8)