Question 1031714: You arrive at a model rocket competition only to discover that you left your small solid fuel engines at home. in a fit of anger, you throw the engine-less rocket straight up in the air. the rocket leaves your hand 6 feet above the ground with an initial throw velocity of 45 feet per second. realizing that you will need the rocket for future competitions, 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?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 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
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