Question

An eagle flies at constant velocity horizontally across the sky, carrying a mouse and releases the mouse while in flight. From the eagle’s perspective, the mouse falls vertically with speed $v_1$ . From an observer on the ground’s perspective, the mouse falls at an angle with speed $v_2$ . What is the speed of the eagle with respect to the observer on the ground ?

• A. $v_1 + v_2$
• B. $v_1 - v_2$
• C. $\sqrt{v^2_1 - v^2_2}$
• D. $\sqrt{v^2_2 + v^2_1}$

Answer 1

Answer: (D)

Let the speed of the eagle is $v$ with respect to ground.
Diagram of velocities, according to the question, where, $v_{1}$ is speed of mouse with respect to the eagle and $v_{2}$ is speed of mouse with respect to the ground.
Thus, we have
$V^{2}=\sqrt{V^{2}+V_{1}^{2}+2VV_{1}\, cos\, 90^{\circ}}$
$\therefore \,V_{2}=\sqrt{V^{2}+V_{1}^{2}}$
$V_{2}^{2}=V^{2}+V_{1}^{2}$
$\therefore \, V=\sqrt{V_{2}^{2}-V_{1}^{2}}$