Question

A eagle launches herself straight up with an initial velocity of 75.0 m/s toward its prey. Theprey is located 250. m above the ground. How fast will the eagle be moving when shereaches her prey?

Answer 1

Given,

The initial velocity of the eagle, u=75.0 m/s

The height at which the prey is located, h=250 m

The acceleration of the eagle is equal to the acceleration due to gravity. And the acceleration due to gravity will be directed downward while the initial velocity of the eagle is directed upwards. Therefore the acceleration of the eagle will be g=-9.8 m/s²

From the equation of motion,

`v^2=u^2+2gh`

Where v is the final velocity of the eagle, i.e., the velocity of the eagle when it catches the prey.

On substituting the known values,

`\begin{gathered} v^2=75.0^2+2*-9.8*250 \\ =725 \\ \Rightarrow v=26.93\text{ m/s} \end{gathered}`

Thus the velocity of the eagle when it reaches her prey is 26.93 m/s