Salmon often jump waterfalls to reach their breeding grounds.
Starting 3.16 m from a waterfall 0.257 m in height, at what minimum speed must a salmon jumping at an angle of 37.9◦ leave the water to continue upstream? The acceleration due to gravity is 9.81 m/s2 .
Answer in units of m/s
To solve this problem, we can break the salmon's motion into horizontal and vertical components. We can then use the kinematic equations to find the minimum speed required.
Horizontal motion:
Let vx be the horizontal component of the salmon's initial velocity.
vx = v * cos(37.9)
The horizontal distance the salmon must travel is 3.16 m. Since the acceleration due to gravity does not affect horizontal motion, the time it takes to travel this distance can be found with the equation:
x = vx * t
3.16 m = v * cos(37.9) * t
Vertical motion:
Similarly, let vy be the vertical component of the salmon's initial velocity.
vy = v * sin(37.9)
The vertical distance the salmon rises is 0.257 m, and the acceleration due to gravity is downward. The equation for the vertical displacement is:
y = vy * t - 0.5 * g * t^2
0.257 m = v * sin(37.9) * t - 0.5 * 9.81 m/s^2 * t^2
Now we have two equations with two unknowns, t and v. We can solve these equations to find the value of v.
Divide the first equation by the second equation to eliminate t:
(3.16 m) / (0.257 m) = (v * cos(37.9) * t) / (v * sin(37.9) * t - 0.5 * 9.81 m/s^2 * t^2)
Cancel out the t term and simplify the equation:
(3.16 m) / (0.257 m) = (cos(37.9)) / (sin(37.9) - (0.5 * 9.81 m/s^2 * t^2)/(v * sin(37.9) * t))
Now get a common denominator and combine the terms:
3.16*cos(37.9) = 0.257*sin(37.9) - 3.16*0.5*9.81*t^2/v*sin(37.9)
Now isolate the v term:
v = (3.16*0.5*9.81*t^2) / (0.257*cos(37.9) - 3.16*sin(37.9))
Finally, plug in the value of t from the horizontal equation into the vertical equation and solve for v:
t = 3.16/(v*cos(37.9))
v = (3.16*0.5*9.81*(3.16/(v*cos(37.9))^2)) / (0.257*cos(37.9)-3.16*sin(37.9))
Solving for v, we find that the minimum speed required for the salmon to jump the waterfall is approximately 4.33 m/s.
So the salmon must leave the water at a minimum speed of 4.33 m/s to continue upstream.
To determine the minimum speed at which a salmon must leave the water to continue upstream, we can use the principle of conservation of energy. At the moment the salmon leaves the water, it has a certain amount of kinetic energy and potential energy. These energies must be sufficient for the fish to overcome the gravitational potential energy and reach the desired height.
The potential energy of the salmon can be calculated using the formula:
Potential Energy = mass x gravity x height
In this case, the mass of the salmon will cancel out because it appears on both sides of the equation. Thus, we can rewrite the formula as:
Potential Energy = gravity x height
The minimum speed required can be calculated using the formula for kinetic energy:
Kinetic Energy = 0.5 x mass x velocity^2
Since we want to find the minimum speed, we can assume that all of the initial kinetic energy will be converted into gravitational potential energy. Therefore, the two energies are equal:
Potential Energy = Kinetic Energy
gravity x height = 0.5 x mass x velocity^2
Rearranging the equation to solve for velocity:
velocity^2 = (2 x gravity x height) / mass
Substituting the known values:
velocity^2 = (2 x 9.81 m/s^2 x 0.257 m) / 3.16 m
velocity^2 = 0.08176 m^2/s^2
Taking the square root to solve for velocity:
velocity = √(0.08176 m^2/s^2)
velocity ≈ 0.286 m/s
Therefore, the salmon must leave the water with a minimum speed of approximately 0.286 m/s to continue upstream.
To find the minimum speed required for the salmon to jump the waterfall, we can use the principles of projectile motion.
The vertical distance the salmon needs to travel is the height of the waterfall, which is 0.257 m. The horizontal distance is 3.16 m.
First, let's find the time it takes for the salmon to reach the top of its trajectory (at the peak of the jump). We can use the equation:
y = v0y * t + (1/2) * a * t^2
where:
- y is the vertical displacement (0.257 m)
- v0y is the vertical component of the initial velocity (unknown)
- a is the acceleration due to gravity (-9.81 m/s^2)
- t is the time
Since the salmon starts and ends at the same vertical position, the displacement is zero at the top of the jump. Therefore, the equation becomes:
0 = v0y * t + (1/2) * (-9.81 m/s^2) * t^2
Simplifying the equation:
(1/2) * (-9.81 m/s^2) * t^2 = v0y * t
Divide both sides by t:
(1/2) * (-9.81 m/s^2) * t = v0y
Next, let's find the horizontal component of the initial velocity (v0x). We can use the formula:
x = v0x * t
where:
- x is the horizontal distance (3.16 m)
- v0x is the horizontal component of the initial velocity (unknown)
- t is the time
Rearranging the equation:
v0x = x / t
We can express t in terms of v0y using the equation derived earlier:
t = v0y / (9.81 m/s^2)
Substituting this value into the equation for v0x:
v0x = x / (v0y / (9.81 m/s^2))
Simplifying the equation:
v0x = (9.81 m/s^2 * x) / v0y
Finally, we can find the minimum speed required for the salmon to continue upstream by calculating the magnitude of the initial velocity (v0):
v0 = sqrt(v0x^2 + v0y^2)
Substituting the previously calculated values:
v0 = sqrt(((9.81 m/s^2 * x) / v0y)^2 + v0y^2)
Now, we can substitute the known values into the equation:
v0 = sqrt(((9.81 m/s^2 * 3.16 m) / v0y)^2 + v0y^2)
To minimize the speed, the salmon should jump at the smallest possible angle (37.9°). We can calculate v0y using the equation:
v0y = v0 * sin(37.9°)
Rearranging the equation:
v0 = v0y / sin(37.9°)
Substituting this value into the equation:
v0 = sqrt(((9.81 m/s^2 * 3.16 m) / (v0 * sin(37.9°)))^2 + (v0 * sin(37.9°))^2)
Simplifying the equation:
v0 = sqrt(((9.81 m/s^2 * 3.16 m) / (v0 * sin(37.9°)))^2 + v0^2 * sin^2(37.9°))
This equation cannot be solved algebraically, so we can use numerical methods such as iteration or approximation to find the value of v0.