Answer
(a) The horizontal component of velocity of the raindrop with respect to the earth is zero.
The horizontal component of velocity of the raindrop with respect to the train is 12.0 m/s to the west.
(b) The magnitude of the velocity of the raindrop with respect to the earth is 20.8 m/s.
The magnitude of the velocity of the raindrop with respect to the train is 24.0 m/s.
Work Step by Step
(a) The raindrop is falling vertically with respect to the earth, so the horizontal component of velocity of the raindrop with respect to the earth is zero.
The train is moving 12.0 m/s to the east, so the horizontal component of velocity of the raindrop with respect to the train is 12.0 m/s to the west.
(b) Let $v_y$ be the vertical speed of the drop with respect to the Earth.
$\frac{12.0~m/s}{v_y} = tan(30.0^{\circ})$
$v_y = \frac{12.0~m/s}{tan(30.0^{\circ})}$
$v_y = 20.8~m/s$
We can find the magnitude of the velocity of the raindrop with respect to the train.
$v = \sqrt{(12.0~m/s)^2+(20.8~m/s)^2}$
$v = 24.0~m/s$
The magnitude of the velocity of the raindrop with respect to the train is 24.0 m/s.
The magnitude of the velocity of the raindrop with respect to the earth is $v_y$, which is 20.8 m/s.
