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:
[tex]v_x = 3.80 m/s[/tex]
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
[tex]v_y = u_y + gt[/tex]
where
[tex]u_y =0[/tex] 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
[tex]v_y = 0 + (9.8)(1.00)=9.8 m/s[/tex]
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:
[tex]v=\sqrt{v_x^2 +v_y^2}=\sqrt{(3.8)^2+(9.8)^2}=10.5 m/s[/tex]
b) [tex]\theta = -68.8^{\circ}[/tex]
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
[tex]\theta = tan^{-1} (\frac{v_y}{v_x})=tan^{-1} (\frac{-9.8}{3.8})=-68.8^{\circ}[/tex]
We have to understand in which direction is this angle measured. In fact, the car is moving forward, so [tex]v_x[/tex] has forward direction (we can say it is positive if we take forward as positive direction).
Also, the ball is moving downward, so [tex]v_y[/tex] 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:
[tex]\theta = -68.8^{\circ}[/tex]
