Question

The beautiful Multnomah Falls in Oregon are approximately 206m high. If the Columbia river is flowing horizontally at 2.90m/s just before going over the falls, what is the overall velocity of the water when it hits the bottom?

Answer 1

Answer:

The overall velocity of the water when it hits the bottom is:

$v_f=63.61\ \frac{m}{s}$

Explanation:

Use the law of conservation of energy.

Call it instant [1] to the moment when the water is just before reaching the falls.

At this moment its height h is 206 meters and its velocity  horizontally $v_i$ is $v_i = 2.90$m/s.

At the instant [1] the water has gravitational power energy $E_g$

$E_g = mgh$

The water also has kinetic energy Ek.

$E_k = 0.5mv_i ^ 2$

Then the Total E1 energy is:

$E_1 = mgh + 0.5mv_i ^ 2$

In the instant [2] the water is within an instant of touching the ground. At this point it only has kinetic energy, since the height h = 0. However at time [2] the water has maximum final velocity $v_f$

So:

$E_2 = 0.5mv_f ^ 2$

As the energy is conserved then $E_1 = E_2$

$mgh + 0.5mv_i ^ 2 = 0.5mv_f ^ 2$

Now we solve for $v_f$.

$gh + 0.5v_i ^ 2 = 0.5v_f ^ 2\\\\9.8(206) + 0.5(2.90) ^ 2 = 0.5v_f 2\\\\v_f^2 = \frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}\\\\v_f = \sqrt{\frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}}\\\\v_f=63.61\ \frac{m}{s}$