A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top?

6.86 m/s

Explanation:

The minimal velocity needed is when we have only vertical motion, then i will think in the problem only in one axis.

I suppose that the only force, in this case, is the gravitational force acting on the fish.

Then the gravitational equation of the fish will be:

a(t) = -9.8m/s^2

For the velocity equation we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial velocity of the fish and is what we want to find.

For the position equation we need to integrate over time again to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + v0*t + p0

p0 is the initial position of the fish, and because he starts one the water, the initial position is p0 = 0 m

Then the equation is:

p(t) = (1/2)*(-9.8 m /s^2)*t^2 + v0*t

p(t) = (-4.9 m/s^2)*t^2 + v0*t

We know that the maximum height is 2.4m

The value of time at which the fish gets his maximum height is when the velocity of the fish is equal to zero, then we first need to solve:

v(t) = (-9.8m/s^2)*t + v0 = 0

      t = v0/9.8m/s^2

Now we replace this in the position equation to get the maxmimum height, which is equal to 2.4m

2.4m = p( v0/9.8m/s^2) =  (1/2)*(-9.8 m /s^2)*(v0/9.8m/s^2)^2 + v0*(v0/9.8m/s^2)

2.4m = (1/2)(-v0)^2(-9.8 m /s^2) + v0^2/(9.8m/s^2))

2.4m = (1 - 1/2)*v0^2/(9.8m/s^2)

2.4m = 0.5*v0^2/(9.8m/s^2)

2.4m/0.5 = v0^2/(9.8m/s^2)

4.8m*(9.8m/s^2) = v0^2

√(4.8m*(9.8m/s^2)) = v0 = 6.86 m/s

weather forecasts are done by collecting data about the current state of the atmosphere mainly the temperature, humidity and wind and using meteorology, to determine the atmosphere that evolves over time. the h and l mean the high and low pressure. the red strip is a warm front. the blue strip is a cold front. the blue and red strip is a stationary front. the purple strip is a occluded front.

explanation:

it is the second option.