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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# An eagle is flying horizontally at a speed of ${\mathbf{3}}{\mathbf{.}}{\mathbf{00}}$m/s when the fish in her talons wiggles loose and falls into the lake ${\mathbf{5}}{\mathbf{.}}{\mathbf{00}}$ m below. Calculate the velocity of the fish relative to the water when it hits the water.

The final velocity of the fish relative to water when it hits the water, vf = $10.3$ m/s.

See the step by step solution

## Step 1: Definition of final velocity

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, vix = $3.00$ m/s.
• Vertical displacement of the fish to the lake, ∆y = $5.00$ m.
• The initial vertical component, viy = 0 m/s.
• Acceleration of the fish in the vertical direction, ay = - $-9.8$m/s2.

## Step 2: Finding final velocity of fish

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

${{V}_{fy}}^{2}={{V}_{i}}^{2}+2{a}_{y}∆Y$

Substitute the values in the above equation, we get

${{V}_{fy}}^{2}=0+2×\left(-9.8\right)×\left(-5.00\right)\phantom{\rule{0ex}{0ex}}{{V}_{fy}}^{2}=98\phantom{\rule{0ex}{0ex}}{V}_{fy}=±\sqrt{98}\phantom{\rule{0ex}{0ex}}=±9.9m/s$

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

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

Now acceleration in the horizontal direction, ax = 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, vf .

The resultant vector

${V}_{f}=\sqrt{{\left(3.00\right)}^{2}+{\left(-9.90\right)}^{2}}\phantom{\rule{0ex}{0ex}}=10.3m/s$

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