Question

An eagle is flying horizontally at a speed of 3.00 m/s when the fish in 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.

Answer 1

Answer:

$v=10.34\frac{m}{s}\\\theta=73.14^\circ$

Explanation:

The velocity of the fish relative to the water when it hits the water is given by its magnitude and its angle:

$v=\sqrt{v_x^2+v_y^2}\\\theta=tan^{-1}(\frac{v_y}{v_x})$

The horizontal speed ($v_x$) of the fish is constant, while vertically we have a free fall ($v_{0y}=0$), using the kinematic equations we can calculate the vertical speed ($v_y$):

$v_y^2=v_{0y}^2+2gh\\v_y^2=2gh$

Now, we calculate the velocity of the fish:

$v=\sqrt{v_x^2+2gh}\\v=\sqrt{(3\frac{m}{s})^2+2(9.8\frac{m}{s^2})(5m)}\\v=10.34\frac{m}{s}\\\theta=tan^{-1}(\frac{\sqrt{2gh}}{v_x})\\\theta=tan^{-1}(\frac{\sqrt{2(9.8\frac{m}{s^2})(5m)}}{3\frac{m}{s}})\\\theta=73.14^\circ$