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A 0.16 kg baseball moving at +21 m/s is slowed to a stop by a catcher who exerts a...
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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.
Posted by krishna-agrawala on October 23, 2009 at 1:41 AM (Answer #1)
High School Teacher
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.
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.
Posted by neela on October 23, 2009 at 1:44 AM (Answer #2)
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