b) Vertical component of the velocity: 16.6 m/s (downward)
a) The final velocity is 16.6 m/s at [tex]57.3^{\circ}[/tex] below the horizontal
Explanation:
b)
The motion of the fish is a projectile motion, which consists of two independent motions:
- A uniform motion along the horizontal direction
- A vertical motion with constant acceleratio
We start by finding the vertical component of the fish velocity at the moment it hits the water.
To do that, we consider only the vertical motion of the fish, which is a free fall motion, so we can apply suvat equations:
[tex]v_y^2 -u_y^2 = 2as[/tex]
where we have:
[tex]v_y[/tex] = vertical velocity of the fish when it enters the water
[tex]u_y = 0[/tex] initial vertical velocity of the fish
[tex]a=g=9.8 m/s^2[/tex] (acceleration of gravity)
s = 14 m vertical displacement
Solving for [tex]v_y[/tex], we find
[tex]v_y = \sqrt{u_y^2+2as}=\sqrt{0+2(9.8)(14)}=16.6 m/s[/tex]
in the downward direction.
a)
We can now find the resultant velocity of the fish as it enters the water. In fact, the horizontal velocity remains constant during the entire fall (as there are no forces acting in this direction), and so it is
[tex]v_x = 9 m/s[/tex]
While we found the vertical velocity to be
[tex]v_y = 14 m/s[/tex]
So, the magnitude of the final velocity is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{9^2+14^2}=16.6 m/s[/tex]
And the direction is given as
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{14}{9})=57.3^{\circ}[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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