Question

A ball is ejected to the right with an unknown horizontal velocity from the top of a pillar that is 50 meters in height. At the exact instant, a carriage moving on rails is also released to the right from the bottom of the pillar. Calculate the velocity with which the carriage should be released so that the ball falls in the carriage after the carriage has traveled a distance of 50 meters on the ground.

Answer 1

15.67 m/s

Step-by-step 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

`S=(1)/(2)gt^2`

where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:

`t=\sqrt{(2S)/(g)}=\sqrt{(2(50 m))/(9.8 m/s^2)}=3.19 s`

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:

`v=(d)/(t)=(50 m)/(3.19 s)=15.67 m/s`