A 0.16 kg baseball moving at +21 m/s is slowed to a stop by a catcher who exerts a constant force of -350 N.How long does it take this force to...

The aceleration of the ball due to the force =force/mass of the ball=-350/0.16 m/s^2

To stop the ball imlies the ball' velocity becomes 0.The relation between the initial and final velocities and the acceleration is

v=u+at, where u is the initial velocity of the ball =21m/s and v isthe final velocity =0 in this case, and ais the acceleration =(350/0.16)m/s^2.t is the time to find out.

t=(v-u)/a=(0-21)/(-350/0.16)=0.0096 secs.

Therefore , the distance s travelled by the ball while stopping is given by: s =(v^2-u^2)/(2a), where u is the final velocity , u is the initial velocity and a is the acceleration. v=0, u=21 and a=-350/0.16. So, s=(0-21^2)/(2*(-350/0.16))= 0.0983m.

Therefore, the ball has traversed a distance of 0.1008 m after the applied force to stop it.