Posted by Tifna on Friday, August 28, 2009 at 3:24am.
How much energy does it take to go 1.5m?
mgh=m*9.8*1.5= 14.7m joules
How much KE does the fish have initially?
1/2 m v^2= 1/2 m (6.26^2)=19.6m
Well, there is plenty of energy to go that high. So, will the fish go that high and still land 1m kupstream?
At the top , vvertical is zero. So how long will the fish take to fall that same distance?
vvertical= vverticalinitial-gt
0=6.26sinTheta-gt
This is the same time it takes to get to the top, so
t= 6.26sinTheta/g
How far does the fish travel hoizontally in that time?
1=6.26cosTheta*t
= 6.26costheta*6.26sinTheta/g
solvefor theta. Remember 2cosAsinA=sin2A
If you get an angle, the fish can do it.
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