Question

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?

Answer 1

a) t = 0.12 s

b) x = 1.082 m

Step-by-step 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