Answer:
The velocity of the fish when it hits the water is:
v = 10.93 m/s and 71.88 ° below the x-direction.
Explanation:
Let's find the velocity of the fish in the y-direction.
[tex]v_{fy}^{2}=v_{iy}^{2}-2gh[/tex]
Here, v(iy) of the fish is zero, and the heigh h = 5.50 m, then the velocity will be:
[tex]v_{fy}^{2}=0-2(9.81)(5.50)[/tex]
[tex]v_{fy}^{2}=-2(9.81)(5.50)[/tex]
[tex]v_{fy}=-10.39 \: m/s[/tex]
Now, we know that the velocity in the x-direction is constant, so we can calculate the velocity of the fish when it hits the water.
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]
[tex]v=\sqrt{3.40^{2}+(-10.39)^{2}}[/tex]
[tex]v=10.93 \: m/s[/tex]
And the direction will be:
[tex]\theta=tan^{-1}(\frac{10.39}{3.40})[/tex]
[tex]\theta=71.88^{\circ}[/tex]
The angle is 71.88 ° belox the x-direction.
I hope it helps you!
