SOLUTION: An object is launched straight up into the air at an initial velocity of 64 feet per second. It is launched from 6 feet off the ground. Its height H, in feet, at t seconds is given

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Question 423045: An object is launched straight up into the air at an initial velocity of 64 feet per second. It is launched from 6 feet off the ground. Its height H, in feet, at t seconds is given by the equation H=-16tē+64t+6. Find all times t that the object is at a height of 54 feet off the ground.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
An object is launched straight up into the air at an initial velocity of 64 feet per second.
It is launched from 6 feet off the ground.
Its height H, in feet, at t seconds is given by the equation H=-16tē+64t+6.
Find all times t that the object is at a height of 54 feet off the ground.
:
H = 54
therefore:
-16tē + 64t + 6 = 54
-16t^2 + 64t + 6 - 54 = 0
-16t^2 + 64t - 48 = 0
Simplify, divide thru by -16; results:
t^2 - 4t + 3 = 0
Factor
(t - 3)(t - 1) = 0
Two solutions
t = 1 sec, 54 ft on the way up
and
t = 3 sec, 54 ft on the way down
:
+graph%28+300%2C+200%2C+-2%2C+6%2C+-50%2C+100%2C+-16x%5E2+%2B+64x+%2B+6%2C+54%29++
green line is at 54 ft