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.
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
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
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.
To find the minimum speed at which a salmon must jump to continue upstream, we can use the principles of projectile motion.
Let's break down the given information and variables:
- Distance downstream (x) = 2.56 m
- Height of the waterfall (h) = 0.332 m
- Angle at which the salmon jumps (θ) = 31.2°
- Acceleration due to gravity (g) = 9.81 m/s^2
To solve this problem, we need to find the minimum initial speed (v0) at which the salmon must jump to reach the height of the waterfall and still have enough vertical component of its velocity to continue moving upstream.
First, let's analyze the vertical motion of the salmon. At the highest point of its trajectory, the vertical displacement will be equal to the height of the waterfall.
Using the vertical motion formula:
h = (v0sinθ)t - (1/2)gt^2
Since the salmon leaves the water and reaches the highest point, its final vertical displacement (h) will be zero. Thus, we can simplify the equation:
0 = (v0sinθ)t - (1/2)gt^2
Next, we solve this equation for time (t):
(1/2)gt^2 = (v0sinθ)t
Rearranging the equation:
(1/2)gt = v0sinθ
Now let's analyze the horizontal motion of the salmon. The horizontal displacement (x) will be equal to the initial horizontal component of velocity (v0cosθ) multiplied by the time of flight (t):
x = v0cosθ * t
Rearranging this equation:
t = x / (v0cosθ)
As we have found expressions for both t and t, we can substitute the expression for t from the second equation into the first equation:
(1/2)g(x / (v0cosθ)) = v0sinθ
Simplifying:
g(x / (2v0cosθ)) = v0sinθ
Multiplying both sides by 2v0cosθ:
gx = 2v0^2sinθcosθ
Now, solving for v0:
v0^2 = (gx) / (2sinθcosθ)
Taking the square root of both sides:
v0 = √((gx) / (2sinθcosθ))
Finally, substituting the given values into the equation and solving for v0:
v0 = √((9.81 m/s^2 * 2.56 m) / (2 * sin(31.2°) * cos(31.2°)))
v0 = √(25.145 / (2 * 0.523599 * 0.848048))
v0 = √(25.145 / 0.884226)
v0 = √28.480 m/s
Therefore, the minimum speed at which the salmon must jump to continue upstream is approximately 5.34 m/s (rounded to two decimal places).