small rocket is launched from a height of 6 feet off of the ground at an initial upward velocity of 64 feet per second. Its height is given by the equation h = -16t^2 + 64t + 6, where h is the height of the rocket in feet and t is the time since liftoff in seconds. At what time is the rocket 54 feet in the air?
A.
1 second
B.
3 seconds
C.
1 second and 3 seconds
D.
none of the above

Question

jhonyy9 · Answer

h=54 feet

- 16t^2 +64t +6 = 54

solve this quadratic for t and will get the correct answer sure

hope helped