Un pitcher lanza una bola rapida horizontalmente con una rapidez de 140 km/h hacia home, que esta a 18.4m de distancia. si los timepo combinados suman 0.350 s, ¿durante cuanto tiempo puede mirar el bateador la bola despues de que sale de la mano del lanzador, antes de hacer el swing? en su recorrido hacia home,¿ que tanto baja la pelota respecto a su linea horizontal original?

a) t = 0.12 s

b) x = 1.082 m

Explanation:

In order to do this, we need to use the expressions for horizontal throwing.

a) First, let's convert the units from km/h to m/s:

140 km/h * 1000 m/km * 1h/3600 s = 38.89 m/s

This is the speed of the ball.

Now, let's calculate the time of the ball to reach home:

V = x/t ---> t = x/V

t = 18.4 / 38.89 = 0.47 s

The combined times is 0.35, therefore, the time for reaction would be the difference of the times:

t = 0.47 - 0.35 = 0.12 s

b) We use the expression of height which is:

h = gt²/2

h = 9.8 * (0.47)²/2

h = 1.082 m

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.