Question 30988
a rocket is launched from atop a 32-foot cliff with an initial velocity of 154 ft/s.
a.write an equation to describe the height of the rocket T seconds after launch
THE EQN. IS 
S=32+UT-GT^2/2...WHERE 
S=DISTANCE TRAVELLED INCLUDING THE INITIAL HT.OR TOTAL HEIGHT ABOVE GROUND......FT.
U=INITIAL VELOCITY...FT/SEC
T= TIME...SEC.
G=ACCELERATION DUE TO GRAVITY...32 FT/SEC.SEC.
SUBSTITUTING...
S=32+154*T - 32*T^2/2 =32+154*T-16*T^2
b.graph the equation to find out how long after the rocket is launched it will hit the ground.
T	0	2	4	6	8	9.8
S	32	276	392	380	240	0.0


{{{ graph( 500, 500, -5, 15, -50, 500, 32+154*x-16*x^2
) }}}
c.estimate your answer to the nearest tenth of a second
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)) }}}
{{{x = (-154 +- sqrt( 154^2-4*-16*32)/(2*-16)) }}}

T=9.8 SEC.......FROM THE QUADRATIC EQN.S=0 AT 9.2 SEC...