An eagle is flying horizontally at a speed of 4.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.20 m below. Calculate the velocity (in m/s) of the fish relative to the water when it hits the water. (Assume that the eagle is flying in the + x-direction and that the + y-direction is up.)

V=11.74m/s, 69.59°

Explanation:

From newtons equation of motion, we know that

V^2 = u^2+2gh

for the rt

For the vertical component of the speed

Vy^2=V0^2 + 2gh

Vy = the final speed in the vertical axis

V0 = initial speed, 0m/s

g = acceleration due to gravity 9.81m/s

h=height/distance between the eagle and the lake

Vy^2 = 2*9.81*6.2

Vy = √121.644

Vy=11.02m/s

The resultant speed will be

V = (Vy^2+Vx^2) ^0.5

V = (11.02^2+4.1^2) ^0.5

V=137.7241^0.5

V=11.74m/s

Direction

Tan^-1 (11.02/4.1)

β=69.59°

V=11.74m/s, 69.59°

Answer = 11.74m/s, b°