Question

An eagle goes straight up with an initial velocity of 75m/s toward its food. Its food is located 250m above the ground. How fast will the eagle be moving when she reaches her food?

Answer 1

The vertical distance covered by the eagle can be given as,

`h=ut+(1)/(2)gt^2`

Plug in the known values,

`\begin{gathered} 250m=(75\text{ m/s)t+}(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}`

The above equation can be further solved as,

`\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}`

Therefore, the time taken by eagle to reach at food is 10.4 s or 4.91 s.