Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2m from a waterfall that is 0.55m tall and jumps at an angle of 32°. What must be the salmon's minimum speed to reach the waterfall?
4.67
The range R of the jump must be R = 4m, if the waterfall were not there. That correspopnds to a distance to maximum height (the waterfall top) of 2 m.
Using the equation for range:
(R/2) = (1/2)*Vo^2*sin(2A)/g = 2 m
Vo^2 = 21.8 m/s
Vo = 4.7 m/s
A is the launch angle. It gets doubled because 2sinA*cosA = sin(2A) appears in the equation for R.
To determine the salmon's minimum speed to reach the waterfall, we need to consider the vertical component of the salmon's initial velocity.
Given:
Distance from the waterfall, d = 2m
Height of the waterfall, h = 0.55m
Angle of the salmon's jump, θ = 32°
To find the minimum speed, we can use the equation for the vertical component of velocity:
v_y = √(2 * g * h)
where:
v_y is the vertical component of velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height of the waterfall
Plugging in the values:
v_y = √(2 * 9.8 * 0.55)
= √(10.78)
≈ 3.28 m/s
Therefore, the salmon's minimum speed to reach the waterfall is approximately 3.28 m/s.
To determine the salmon's minimum speed to reach the waterfall, we can use the principles of projectile motion.
Let's break down the problem into components: horizontal (x-direction) and vertical (y-direction).
In the y-direction:
1. The initial vertical velocity (Vy) is the salmon's minimum speed since it requires exactly the minimum speed to reach the waterfall.
2. The initial vertical position (y0) is 2 meters.
3. The vertical displacement (Δy) is the height of the waterfall, which is 0.55 meters.
4. The angle of the jump (θ) is given as 32°.
5. The vertical acceleration (ay) is the acceleration due to gravity, which is approximately -9.8 m/s² (negative because it acts downward).
Using the kinematic equation for vertical motion:
Δy = Vy₀ * t + (1/2) * ay * t²
Since the salmon must reach the waterfall, the vertical displacement Δy is equal to the height of the waterfall. Plugging in the known values:
0.55 = Vy * t + (1/2) * (-9.8) * t²
In the x-direction:
1. The initial horizontal velocity (Vx) is the horizontal component of the salmon's speed, which we need to find.
2. The distance from the waterfall (x0) is unknown.
3. The horizontal acceleration (ax) is 0 since there is no horizontal force acting on the salmon.
For projectile motion, the time taken to reach the highest point of the trajectory (t) is equal to the time taken to reach the waterfall on its way down.
Using the equation for horizontal motion:
x = Vx * t
Since we are looking for the minimum speed, we want to minimize the time taken. This occurs when the salmon reaches the waterfall at the peak of its trajectory, meaning t is the time taken to reach the highest point.
Now, we have two equations with two unknowns (Vy and t). We can solve them simultaneously to find the minimum speed required for the salmon.
Solving the vertical equation for t:
t = (Vy - sqrt(Vy² - 4 * (-4.9) * 2)) / (-9.8)
Substituting this expression for t in the horizontal equation:
x = Vx * (Vy - sqrt(Vy² - 4 * (-4.9) * 2)) / (-9.8)
Since x is the distance from the waterfall (given as 2m), we can solve this equation for Vx:
Vx = -x * (-9.8) / (Vy - sqrt(Vy² - 4 * (-4.9) * 2))
Finally, substituting the given angle (θ = 32°) and rearranging the terms, we can solve for Vy:
Vy = sqrt(x * (-9.8) / (2 * (1 - cos(θ))))
Now, substituting the known values:
Vy = sqrt(2 * (-9.8) * (-0.55) / (2 * (1 - cos(32°))))
Simplifying this expression gives the minimum speed (Vy) for the salmon to reach the waterfall.