An eagle flies at constant velocity horizontally across the sky, carrying a mouse and releases the mouse while in flight. From the eagle?? perspective, the mouse falls vertically with speed $v_1$ . From an observer on the ground?? 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 ?
- $v_1 + v_2$
- $v_1 - v_2$
- $\sqrt{v^2_1 - v^2_2}$
- $\sqrt{v^2_2 + v^2_1}$
The Correct Option is D
Solution and Explanation
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
$V^{2}=\sqrt{V^{2}+V_{1}^{2}+2VV_{1}\, cos\, 90^{\circ}}$
$\therefore \,V_{2}=\sqrt{V^{2}+V_{1}^{2}}$
$V_{2}^{2}=V^{2}+V_{1}^{2}$
$\therefore \, V=\sqrt{V_{2}^{2}-V_{1}^{2}}$
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Concepts Used:
Relative Velocity
The velocity with which one object moves with respect to another object is the relative velocity of an object with respect to another. By relative velocity, we can further understand the time rate of change in the relative position of one object with respect to another.
It is generally used to describe the motion of moving boats through water, airplanes in the wind, etc. According to the person as an observer inside the object, we can compute the velocity very easily.
The velocity of the body A – the velocity of the body B = The relative velocity of A with respect to B
V_{AB} = V_{A} – V_{B}
Where,
The relative velocity of the body A with respect to the body B = V_{AB}
The velocity of the body A = V_{A}
The velocity of body B = V_{B}