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Algebra question...Gravity formula?
A rock is thrown upward with a velocity of 17 meters per second from the top of a 25 meter high liff, and it misses the cliff on the way back down. When will the rock be 8 meters from the water, below?
1 Answer
- ZardozLv 78 years agoFavorite Answer
The hight of a thrown rock is h = vt & at²/2.
Your rock has a hight of interest of h = 8 m - 25 m = -17 m
-17 = (17)t & (-9.81/2)t²; ergo
-4.905t² & 17t & 17 = 0; and
t = {-17 ± √(17² - 4(-4.905)(17))}/2(-4.905)
t = {-17 ± √(289 & 333.54)}/-9.81
t = (-17 ± √622.54)/-9.81; so,
t = (-17 & 24.95)/-9.81; or,
t = (-17 - 24.95)/-9.81
t = 7.95/-9.81 = -0.81 seconds, aka; before you threw it; or,
t = -41.95/-9.81 = 4.28 seconds after you threw it.
Edit: I don't know why but my plus signs ( ) are not rendering. Okay, I've replaced them with an ampersand (&). Let's see if that renders.
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Source(s): [n] = 10ⁿ
