Question

An eagle carry a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1m. The fish travels7.9 m horizontally before hitting the water. What is the velocity of the eagle?

Answer 1

Answer:

Velocity of the eagle = 7.08 m/s

Explanation:

  We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  Vertical motion of fish:

      Initial velocity = 0 m/s , acceleration = acceleration due to gravity = $9.8m/s^2$, displacement = 6.1 m we need to find time.

     $6.1= 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2=6.1\\ \\ t=1.116 seconds$

 Horizontal motion of fish:

      Acceleration = acceleration due to gravity = $0m/s^2$, displacement = 7.9 m , time = 1.116 seconds we need to find Initial velocity (u).

      $7.9= u*1.116+\frac{1}{2} *0*1.116^2\\ \\ u=7.08m/s$

    So, velocity of the eagle = 7.08 m/s