Answer:
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
Answer:
The speed of fish at the time of hitting the surface is 9.95 m/s.
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
[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]
The net velocity with which the fish strikes to the water is
[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]
