An eagle is flying horizontally at a speed of 3.00 m/s when the fishin her talons wiggles loose and falls into the lake 5.00m below. calculate the velocity of the fish relative to the water when it hits the water.
Vf = √[Vi² + 2*g*y] = 10.34 m/s
Θ = arccos[Vi/Vf] = 73.1° below horizontal
To determine the magnitude to the velocity of fish when it hits the water, use the following equation.
v = √(Final horizontal velocity^2 + Final vertical velocity^2)
The final horizontal velocity is the same as the eagle’s initial speed.
v = √(9 + Final vertical velocity^2)
Since we don’t know the time the fish is falling, let’s use the following equation to determine the fish’s vertical velocity when it hits the water.
vf^2 = vi^2 + 2 * a * d, vi = 0, a = 9.8, d = 5
vf^2 = 2 * 9.8 * 5 = 98
Let’s put this number into the equation and solve for v.
v = √(9 + 98)
The magnitude of the velocity is approximately 10.34 m/s. Since the question asks for the final velocity, we need to use the following equation to determine the angle with respect to the water. Use the following equation.
Tan θ = Vertical/Horizontal = 3/√107
The angle is approximately 16.2˚ above horizontal.
Vertical V after 5m. fall = sqrt.(2gh) = 9.9m/sec.
V at water = sqrt.(9.9^2 + 3^2)10.34m/sec.
Direction at water = arctan (9.9/3) = 73.14 degrees below horizontal.