Let V be the initial velocity leaving the water. The vertical component of the velocity, V sin 45, must be enough to let the salmon reach a height of h = 3.58 m
Maximum altitude h is reached when
V sin 45 = g T
The maximum height reached is then
h = (1/2) V sin 45 T
= (1/2) [V sin 45]^2/g = (1/4)V^2/g
V =sqrt (4 g h) = 2 sqrt (g h)
Salmon, swimming up the Fraser river to their spawning grounds, leap over all sorts of obstacles. The unofficial salmon-altitude record is an amazing 3.58 m jump. Assuming the fish took off at 45.0°, what was its speed on emerging from the water? Ignore friction.
This problem is driving me CRAZY. I've tried this many times, but I have a feeling I'm making it a lot harder than it really is. Could someone tell me which equations I need to use?
3 answers
THANK YOU!
v = 11.8 m/s
That is faster than the fastest human can run horizontally (10 m/s)!
That is faster than the fastest human can run horizontally (10 m/s)!