Physics, 17.12.2019 18:31 kharmaculpepper

Achinook (king) salmon (genus oncorynchus) can jump out of water with a speed of 6.75 m / s . if the salmon is in a stream with water speed equal to 1.65 m / s , how high in the air can the fish jump if it leaves the water traveling vertically upwards relative to the earth?

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Answer from: hvvhvc

The maximum height that the fish can jump is 2.19 m.

Explanation:

Hi there!

Please, see the attached figure for a better understanding of the problem.

The motion of the salmon is a parabolic one because when it jumps, it already has a horizontal velocity (see figure).

The position and velocity vectors of the salmon at a time t, can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position of the salmon at time t.

x0 = initial horizotal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration of gravity.

Looking at the figure, notice that at the maximum height, the vertical velocity is zero (because the velocity vector is horizontal). Using the equation of the vertical component of the velocity, we can obtain the time at which the salmon is at its maximum height:

vy = v0y + g · t

To find the initial vertical velocity, v0y, let´s look at the figure. Notice that the initial velocity is the hypotenuse of the triangle formed with the horizontal velocity and the vertical velocity. Then:

v0² = v0x² + v0y²

Solving for v0y:

v0y = √(v0² - v0x²)

v0y = √((6.75 m/s)² - (1.65 m/s)²)

v0y = 6.55 m/s

Now, using the equation of the vertical component of the velocity at the maximum height (vy = 0):

vy = v0y + g · t

0 = 6.55 m/s + (-9.8 m/s²) · t

-6.55 m/s / -9.8 m/s² = t

t = 0.67 s

Now, using the equation of the vertical position at t = 0.67 s, we can find the maximum height:

y = y0 + v0y · t + 1/2 · g · t²

y = 0 m + 6.55 m/s · 0.67 s + 1/2 · (-9.8 m/s²) · (0.67 s)²

y = 2.19 m

The maximum height that the fish can jump is 2.19 m.

Achinook (king) salmon (genus oncorynchus) can jump out of water with a speed of 6.75 m / s . if the

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

It is d, all the other ones say incorrect answer .

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