Salmon often jump waterfalls to reach their
breeding grounds.
Starting 1.77 m from a waterfall 0.253 m
in height, at what minimum speed must a
salmon jumping at an angle of 44.8� leave the
water to continue upstream? The acceleration
due to gravity is 9.81 m/s2 .
Answer in units of m/s
7.5
To calculate the minimum speed at which a salmon must leave the water to continue upstream, we need to consider the energy conservation principle.
The initial mechanical energy of the salmon can be expressed as the sum of its kinetic energy and potential energy. At the start, the salmon only has potential energy. At the peak of the jump, it only has kinetic energy.
The potential energy at the start is given by the formula:
Potential Energy = mass * gravity * height
In this case, the mass of the salmon is not given, but it cancels out when calculating the minimum speed. So we can ignore it for now.
Potential Energy = gravity * height
Potential Energy = 9.81 m/s^2 * 0.253 m
Potential Energy = 2.481 m^2/s^2
At the peak of the jump, all of this potential energy is converted into kinetic energy. The kinetic energy can be expressed as:
Kinetic Energy = (1/2) * mass * velocity^2
Again, we can ignore the mass since it cancels out when calculating the minimum speed.
(1/2) * velocity^2 = 2.481 m^2/s^2
velocity^2 = (2 * 2.481 m^2/s^2)
velocity^2 = 4.962 m^2/s^2
To find the minimum speed, we need to find the square root of this value:
velocity = sqrt(4.962 m^2/s^2)
velocity ≈ 2.228 m/s
Therefore, the minimum speed at which the salmon must leave the water to continue upstream is approximately 2.228 m/s.
To find the minimum speed at which the salmon must leave the water to continue upstream, we can use the principles of projectile motion.
First, let's divide the motion into horizontal and vertical components. Let vx be the horizontal velocity and vy be the vertical velocity of the salmon.
The horizontal motion is unaffected by gravity, so the horizontal velocity, vx, remains constant throughout the motion.
The vertical motion is affected by gravity, so the vertical velocity, vy, changes over time. We can calculate the vertical velocity using the following equation:
vy = u * sin(θ) - g * t
where u is the initial vertical velocity, θ is the launch angle (44.8 degrees), g is the acceleration due to gravity (9.81 m/s^2), and t is the time of flight.
At its peak, the vertical velocity vy will be zero. So, we can write the equation:
0 = u * sin(θ) - g * t_peak
where t_peak is the time it takes for the salmon to reach its peak height.
Since the vertical displacement at this point is equal to the height of the waterfall (0.253 m), we can write another equation:
0.253 = u * sin(θ) * t_peak - (1/2) * g * t_peak^2
Now, let's solve these equations simultaneously to find u (the initial vertical velocity):
To eliminate t_peak, we can solve the first equation for t_peak and substitute it into the second equation:
t_peak = u * sin(θ) / g
0.253 = (u * sin(θ) * u * sin(θ) / g) - (1/2) * g * (u * sin(θ) / g)^2
Simplifying,
0.253 = (u^2 * sin^2(θ) / g) - (1/2) * (u^2 * sin^2(θ) / g)
Now, let's solve for u^2:
2 * 0.253 = 2 * u^2 * sin^2(θ) / g - u^2 * sin^2(θ) / g
0.506 = u^2 * sin^2(θ) / g
u^2 = 0.506 * g / sin^2(θ)
Finally, we can take the square root of both sides to find u:
u = sqrt(0.506 * g / sin^2(θ))
Now, plug in the values:
g = 9.81 m/s^2
θ = 44.8 degrees
u = sqrt(0.506 * 9.81 / sin^2(44.8))
u ≈ 4.18 m/s
Therefore, the minimum speed at which the salmon must leave the water to continue upstream is approximately 4.18 m/s.