The vertical distance covered by the eagle can be given as,
[tex]h=ut+\frac{1}{2}gt^2[/tex]Plug in the known values,
[tex]\begin{gathered} 250m=(75\text{ m/s)t+}\frac{1}{2}(-9.8m/s^2)t^2 \\ -(4.90ms^{-2})t^2+(75\text{ m/s)t-250m=0} \\ (4.90ms^{-2})t^2-(75\text{ m/s)t+250 m=0} \end{gathered}[/tex]The above equation can be further solved as,
[tex]\begin{gathered} t=\frac{75\text{ m/s}\pm\sqrt[]{(75m/s)^2-4(4.90ms^{-2})(250\text{ m)}}}{2(4.90ms^{-2})^{}} \\ =\frac{75\text{ m/s}\pm26.9\text{ m/s}}{9.80m/s^2} \\ =10.4\text{ s, }4.91\text{ s} \end{gathered}[/tex]Therefore, the time taken by eagle to reach at food is 10.4 s or 4.91 s.
