Answer:
15.67 m/s
Explanation:
The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.
Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by
[tex]S=\frac{1}{2}gt^2[/tex]
where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:
[tex]t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(50 m)}{9.8 m/s^2}}=3.19 s[/tex]
Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:
[tex]v=\frac{d}{t}=\frac{50 m}{3.19 s}=15.67 m/s[/tex]
