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 stop the ball? Answer in units of s.

How far does the ball travel before stopping? Answer in units of m.

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we know acceleration = force/mass, or a = f/m

Therefore acceleration of ball = -350/0.16

Pleas note we have taken force and acceleration as negative because it acts in direction opposite to initial velocity of the ball.

The time taken to stop the ball is given by formula:

time = initial velocity/acceleration, or t = u/a = 21/(350/0.16) = 0.0096 seconds

the distance travelled by the ball = (initial velocity/2)*t = 21*0.0096 = 0.1008 m.

Answer:It takes 0.0096 s to stop the ball, and the ball travels 0.1008 m before stopping.

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.

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