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. $\sqrt{v_1^2−v_2^2}$
• D. $\sqrt{v_2^2+v_1^2}$

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
$V^2=\sqrt{V^2+V_1^2+2V{V}_{1}cos90^{∘}}$
$∴V_2=\sqrt{V^2+V_1^2}$
$V_2^2=V^2+V_1^2$
$∴V=\sqrt{V_2^2−V_1^2}$