91Ó°ÊÓ

When gravity first exerts a force on an item, its initial velocity indicates how fast it travels. The final velocity, on the other hand, is a vector number that measures a moving body's speed and direction after it has reached its maximum acceleration.

Given data:

• Speed of the eagle in the x-direction, v_ix = $3.00$m/s.
• Vertical displacement of the fish to the lake, ∆y =$5.00$ m.
• The initial vertical component, v_iy = 0 m/s.
• Acceleration of the fish in the vertical direction, a_y = -$-9.8$m/s².

To find the final velocity in the negative y-direction, we have the kinematic equation

${V_{fy}}^2={V_i}^2+2a_y∆Y$

Substitute the values in the above equation, we get

$$\begin{gathered} {V_{fy}}^2=0+2×(-9.8)×(-5.00) \\ {V_{fy}}^2=98 \\ {V}_{fy}=Â\pm \sqrt{98} \\ =Â\pm 9.9m/s \end{gathered}$$

We take only the negative value because of this velocity in the negative y-direction.

So,${\mathit{V}}_{\mathbf{f}\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{9}\mathit{m}\mathbf{/}\mathit{s}$

Now acceleration in the horizontal direction, a_x= 0 m/s²{since, g = 0 m/s²}.

So the final velocity of the fish in the x-direction is equal to the initial velocity of the fish in the x-direction$3.00$= m/s.

That is, $V_{fx}=V_{ix}=3.00m/s$

So we have, $V_{fx}=3.00m/s$and $V_{fy}=-9.90m/s$

These two velocities are the two components of the final velocity of the fish,v_f .

The resultant vector

$$\begin{gathered} V_f=\sqrt{{(3.00)}^2+{(-9.90)}^2} \\ =10.3m/s \end{gathered}$$

The final velocity of the fish is 10.3 m/s.