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Q40PE

Expert-verifiedFound in: Page 124

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, *v*_{f} = $10.3$ m/s.

**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/s^{2}.

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

${{V}_{fy}}^{2}={{V}_{i}}^{2}+2{a}_{y}\u2206Y$

Substitute the values in the above equation, we get

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

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/s^{2 }{since, g = 0 m/s^{2 }}.

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

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

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

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