Question

An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.70 m below. Calculate the velocity (in m/s) of the fish just before it hits the water. (Assume that the eagle is flying in the x direction and that the y direction is up.)

Answer 1

The speed of fish at the time of hitting the surface is 9.95 m/s.

Step-by-step explanation:

Horizontal speed, u = 2.6 m/s

height, h = 4.7 m

Let the vertical velocity at the time of hitting to water is v.

Use third equation of motion

`v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 * 9.8* 4.7\\\\v = 9.6 m/s`

The net velocity with which the fish strikes to the water is

`v' = √(9.6^2 + 2.6^2 )\\\\v' = 9.95 m/s`

Answer 2

Step-by-step explanation:

The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .

v² = u² + 2 g H .

v² = 0 + 2 x 9.8 x 4.7 m

= 92.12

v = 9.6 m /s

The fish's final velocity will have two components

vertical component = 9.6 m/s downwards

Horizontal component = 2.6 m /s .

Resultant velocity = √ ( 9.6² + 2.6² )

= √ ( 92.16 + 6.76 )

= 9.9 m /s