Question 1031714
 the rocket leaves your hand 6 feet above the ground with an initial throw velocity of 45 feet per second.
 you catch the rocket when it falls back to a height of 5 feet.
 for how many seconds was the rocket in the air?
 what formula gives us this information?
:
-16t^2 + 45t + 6 = 5, where
t = time in seconds
Initial height = 6
Final height = 5
then
-16t^2 + 45t + 6 - 5 = 0
-16t^2 + 45t  + 1 = 0
Using the quadratic formula; a=-16; b=45; c=1, I got a positive solution of:
t = 2.83455 seconds in the air