Salmon, swimming up the Fraser river to their spawning grounds, leap over all sorts of obstacles. The unofficial salmon-altitude record is an amazing 3.54 m jump (vertical distance). Assuming the fish took off at 45.0 degrees, what was its speed on emerging from the water ? Ignore friction.
V^2 = Vo + 2g*h.
V^2 = 0 + 19.6*3.54 = 69.4
V = 8.33 m/s. = Yo = Ver. component of
Vo.
Vo = Yo/sinA = 8.33/sin45=11.78 m/s@45o.
To find the speed of the fish on emerging from the water, we can use the principles of projectile motion. We will consider the vertical and horizontal components of the fish's motion separately.
Let's start with the vertical motion. We know that the fish jumped to a height of 3.54 meters. We can use the equation of motion for vertical displacement to find the initial vertical velocity (v₀) of the fish when it left the water:
Δy = (v₀ * t) + (0.5 * a * t²)
Since the fish starts from rest vertically (it leaves the water without any vertical velocity), the equation simplifies to:
Δy = 0.5 * a * t²
Where:
Δy = vertical displacement = 3.54 m
a = acceleration due to gravity = 9.8 m/s²
t = time of flight
We can rearrange the equation and solve for t:
t = √(2 * Δy / a)
Plugging in the values, we have:
t = √(2 * 3.54 / 9.8)
t ≈ 0.84 seconds
Now, let's move on to the horizontal motion. Given that the fish took off at an angle of 45.0 degrees, we know that the horizontal component of its velocity (v₀x) is equal to its vertical initial velocity (v₀y). We can find this velocity using the equation:
v₀y = v₀ * sin(θ)
Where:
v₀y = initial vertical velocity
v₀ = initial velocity
θ = launch angle = 45.0 degrees
Since v₀y is the same as the vertical component of the final velocity (vfy) when the fish emerges from the water, we can rewrite it as:
vfy = v₀ * sin(θ)
To find the horizontal velocity (vfx), we use the equation:
vfx = v₀ * cos(θ)
Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the flight.
Now, to calculate the speed (v) on emerging from the water, we combine the horizontal and vertical components of the velocity using the Pythagorean theorem:
v = √(vfx² + vfy²)
Substituting the values:
v = √((v₀ * cos(θ))² + (v₀ * sin(θ))²)
Now we can plug in the known values:
θ = 45.0 degrees
v₀ = unknown
v = unknown
However, we do not have enough information to determine the exact value of v₀ and thus v. It is necessary to know either the initial speed or the time taken for the fish to leave the water in order to calculate v.