Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. the car moves horizontally with a constant speed of 3.80 m/s relative to observer 2 standing on the sidewalk.
a) what is the speed of the ball 1.00 s after it is released, as measured by observer 2?
b) what is the direction of travel of the ball 1.00 s after it is released, as measured relative to the horizontal by observer 2?

a) 10.5 m/s

While for observer 1, in motion with the car, the ball falls down straight vertically, according to observer 2, which is at rest, the ball is also moving with a horizontal speed of:

As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

where

is the initial vertical speed

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Therefore, after t = 1.00 s, the vertical speed is

And so the speed of the ball, as observed by observer 2 at rest, is given by the resultant of the horizontal and vertical speed:

b)

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

vapour or gas assuming atmospheriv pressure

mass and volume are the answers you are looking for