Answer:
The overall velocity of the water when it hits the bottom is:
[tex]v_f=63.61\ \frac{m}{s}[/tex]
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 [tex]v_i[/tex] is [tex]v_i = 2.90[/tex]m/s.
At the instant [1] the water has gravitational power energy [tex]E_g[/tex]
[tex]E_g = mgh[/tex]
The water also has kinetic energy Ek.
[tex]E_k = 0.5mv_i ^ 2[/tex]
Then the Total E1 energy is:
[tex]E_1 = mgh + 0.5mv_i ^ 2[/tex]
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 [tex]v_f[/tex]
So:
[tex]E_2 = 0.5mv_f ^ 2[/tex]
As the energy is conserved then [tex]E_1 = E_2[/tex]
[tex]mgh + 0.5mv_i ^ 2 = 0.5mv_f ^ 2[/tex]
Now we solve for [tex]v_f[/tex].
[tex]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}[/tex]
