On a rainy day, rain drops fall vertically with speed \(v\). An observer is moving towards east with speed \(v\), while another observer moves towards north with the same speed \(v\). What is the angle between apparent rain directions observed by them?
\(60^\circ\)
Take east as the x-axis, north as the y-axis and vertically up as the z-axis. The rain's velocity is \(\vec{v}_{rain} = (0, 0, -v)\).
Velocity of rain relative to the observer moving east: \(\vec{v}_1 = (0,0,-v) - (v,0,0) = (-v, 0, -v)\), which makes \(45^\circ\) with the vertical, tilted towards the west.
Velocity of rain relative to the observer moving north: \(\vec{v}_2 = (0,0,-v) - (0,v,0) = (0,-v,-v)\), tilted \(45^\circ\) from the vertical, towards the south.
Angle between the two apparent directions: \(\cos\theta = \dfrac{\vec{v}_1\cdot\vec{v}_2}{|\vec{v}_1||\vec{v}_2|} = \dfrac{v^2}{(\sqrt{2}v)(\sqrt{2}v)} = \dfrac{1}{2}\), so \(\theta = 60^\circ\). Hence option (d) is correct.
Important Questions from Relative Velocity
Rain is falling vertically on the ground at speed 5√3 m/s. If a man walks towards the East with speed 5 m/s, he will feel the rain falling at what angle to the vertical ?
When a stone thrown directly upwards reaches the top, it:
A. Velocity and acceleration are zero.
B. The velocity is zero and the acceleration is about 10 m/s 2.
C. Velocity is about 10 m/s and acceleration is zero.
D. The velocity is about 10 m/s and the acceleration remains the same.
When two bodies move uniformly towards each other, the distance decrease by 6 m/s. If both the bodies moves (as above) in the same direction with the same speed, the distance between them increases by 4 m/s. Then the speed of the two bodies are :
