Salmon often jump waterfalls to reach their
breeding grounds.
Starting downstream, 2.56 m away from a
waterfall 0.332 m in height, at what minimum
speed must a salmon jumping at an angle of
31.2
◦
leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s
2
.
Answer in units of m/s.
5 answers
Ineed help
call that speed u
If there were no gravity, how high above the wall would he pass (height Hi)?
tan 31.2 = Hi/2.56
Hi = 1.55 meters
so in getting to the wall the salmon fell down from 1.55 to .332
or he fell 1.22 meters
how long does that take?
1.22 = (1/2) g t^2 = 4.9 t^2
so t^2 = .249
t = .499 seconds in the air
so it took him .499 seconds to travel 2.56 meters HORIZONTAL
u = 2.56 / .499 = 5.13 m/s
speed = u/cos 31.2 = 5.13/.855 = 6 meters/s
If there were no gravity, how high above the wall would he pass (height Hi)?
tan 31.2 = Hi/2.56
Hi = 1.55 meters
so in getting to the wall the salmon fell down from 1.55 to .332
or he fell 1.22 meters
how long does that take?
1.22 = (1/2) g t^2 = 4.9 t^2
so t^2 = .249
t = .499 seconds in the air
so it took him .499 seconds to travel 2.56 meters HORIZONTAL
u = 2.56 / .499 = 5.13 m/s
speed = u/cos 31.2 = 5.13/.855 = 6 meters/s
Just plug your numbers into the standard trajectory equation
y = tanθ x - g/(2(v cosθ)^2) x^2
and solve for v when y(2.56) = 0.332
y = tanθ x - g/(2(v cosθ)^2) x^2
and solve for v when y(2.56) = 0.332
I mean call horizontal speed u :)
d = Vo^2*sin(2A)/g = 2.56.
Vo^2*sin(62.4)/9.8 = 2.56.
Vo^2*0.0903 = 2.56,
Vo = 5.32 m/s.
Vo^2*sin(62.4)/9.8 = 2.56.
Vo^2*0.0903 = 2.56,
Vo = 5.32 m/s.
6 answers