Asked by Lindsay
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?
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?
Answers
Answered by
drwls
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)
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)
Answered by
Lindsay
THANK YOU!
Answered by
drwls
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)!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.