A 480 g peregrine falcon reaches a speed of 69 m/s in a vertical dive called a stoop. If we assume that the falcon speeds up under the influence of gravity only, what is the minimum height of the dive needed to achieve this speed?

We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy. We will take the point where the Falcon reaches the speed of 69 (m/s), as the point where the potential energy is zero, i.e. it will be the reference point.

At the reference point all potential energy has been transformed into kinetic energy, therefore the kinetic energy can be calculated.

[tex]E_{k}=0.5*m*v^{2} \\ where:\\v = velocity = 69 [m/s]\\m = mass = 480[g] = 0.480[kg]\\E_{k} = kinetic energy [J]\\E_{k} =0.5*0.48*(69)^{2} \\E_{k} =1142.64[J][/tex]

Now we can calculate the elevation with respect to the reference point using the definition of the potential energy.

[tex]E_{p}=m*g*h\\ E_{p}=E_{k} \\therefore\\h= E_{p}/(m*g)\\h= 1142.64/(.48*9.81)\\h=242[m][/tex]

aksnkj aksnkj · Answer 2 · 2021-10-08T11:50:53+02:00

Answer:

The minimum height of the dive needed to achieve the given speed [tex]v=69\rm \; m/s[/tex] is 242.9 m.

Explanation:

Given information:

The mass of peregrine falcon is, [tex]m=480\rm \; g[/tex].

The final speed reached by the peregrine falcon in a vertical dive is, [tex]v=69\rm \; m/s[/tex].

It is given that the falcon is diving vertically downward. It can be compared with the same situation as the freely falling object under the effect of gravity only. So, the initial velocity of the falcon will be [tex]u=0\rm \; m/s[/tex] as the motion starts with rest.

The value of gravitational acceleration or gravity is, [tex]g=9.8\rm \; m/s^2[/tex].

Now, using the third equation of motion, the minimum height [tex]h[/tex] required to the final speed will be,

[tex]v^2-u^2=2gh\\69^2-0^2=2\times9.8\times h\\h=242.9\rm \; m[/tex]

Therefore, the minimum height of the dive needed to achieve the given speed is 242.9 m.

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