Question

An eagle is flying horizontally at a speed of 3.1 m/s when the fish in her talons wiggles loose and falls into the lake 5.8 m below. A. Calculate the magnitude of the velocity of the fish relative to the water when it hits the water in m/s.
B. Calculate the angle, in degrees, by which the fish's velocity is directed below the horizontal when the fish hits the water.

Answer 1

Answer:11.101 m/s

Explanation:

Given

Velocity of eagle=3.1 m/s

Eagle is at height of 5.8 m

Fish is dropped with a horizontal velocity of 3.1 m/s

time taken by fish to reach lake is

$h=ut+\frac{gt^2}{2}$

here u=0 (vertical velocity)

$5.8=\frac{9.8\times t^2}{2}$

t=1.087 s

Vertical velocity acquired by fish is v=u+at

$v=0+9.8\times 1.087=10.66 m/s$

Horizontal velocity=3.1 m/s

Therefore net velocity$=\sqrt{10.66^2+3.1^2}=11.101 m/s$

$tan\theta =\frac{v_y}{v_x}=\frac{10.66}{3.1}=3.43$

$\theta =73.74^{\circ}$ below horizontal