SOLUTION: 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 b.graph the

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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
b.graph the equation to find out how long after the rocket is launched it will hit the ground.
c.estimate your answer to the nearest tenth of a second
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
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%28+500%2C+500%2C+-5%2C+15%2C+-50%2C+500%2C+32%2B154%2Ax-16%2Ax%5E2%0D%0A%29+
c.estimate your answer to the nearest tenth of a second
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29%29+
x+=+%28-154+%2B-+sqrt%28+154%5E2-4%2A-16%2A32%29%2F%282%2A-16%29%29+
T=9.8 SEC.......FROM THE QUADRATIC EQN.S=0 AT 9.2 SEC...