hey everyone...help me out with this one..its a picture

Question

anonymous · Answer

attachment

anonymous · Answer

??

amistre64 · Answer

its a question about seperating the velocity vectors into components that are attributed to fight against gravity and the distance it covers

amistre64 · Answer

-g t^2 +Vj sin(a) t + Hj = 0 will give us the time that the water stays in the air.

Vj cos(a) t = d  , this gives us the distance it covers in the time alotted.

amistre64 · Answer

since: t = d/(Vj cos(a)), we can sub it above to get a time value

-g (d/(Vj cos(a)))^2 + Vj sin(a) * d/(Vj cos(a)) .... ugh, how are you spose to answer?  they give no solid values for anything

amistre64 · Answer

H might end up not even hitting the building

anonymous · Answer

i know lol

anonymous · Answer

it looks like they want an equation

amistre64 · Answer

i cant see a good way to get to it ...

anonymous · Answer

lol
least im not the only one

amistre64 · Answer

there is prolly a physics formula that can be used ... somewhere on the internet

anonymous · Answer

alright ill look
thanks anyway

amistre64 · Answer

stuff like projectile motion: http://electron9.phys.utk.edu/phys135d/modules/m3/Projectile%20motion.htm

amistre64 · Answer

http://www.mrfizix.com/home/projectilemotion.htm