Abaseball pitcher throws a ball horizontally at a speed of 34.0 m/s. a catcher is 18.6 m away from the pitcher. find the magnitude, in meters, of the vertical distance that the ball drops as it moves from the pitcher to the catcher. ignore air resistance.

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

Where,

a = acceleration

t = time

= Initial velocity

= initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

x = Displacement

t = time

With the data we have we can find the time as well

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.