Playing shortstop, you pick up a ground ball and throw it to second base. the ball is thrown horizontally, with a speed of 25 m/s , directly toward point a. when the ball reaches the second baseman 0.42 s later, it is caught at point b. how far were you from the second baseman?

Given:

Initial velocity(u): 25 m/s

Final velocity(v): 0 m/s

Time : 0.42 secs

Now we know that

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

Substituting the given values we get

v=u+at

0=25+a x 0.42

a= -59.52 m/s^2

Now we also know that

v^2= u^2 + 2as

Where v is the final velocity measured in m/s

u is the initial velocity measured in m/s

a is the acceleration measured in m/s^2

s is the displacement/distance traveled measured in m.

Now substituting the given values in the above formula we get

v^2= u^2 + 2 as

0= 25^2 - (2 x 59.52 x s)

s = 5.25 m

Thus the distance traveled is 5.25 m.

spare some change and give it to the poor

explanation:

plz : ((