Final answer:
The baseball player's speed immediately after stopping the ball is approximately 0.044 m/s.
Step-by-step explanation:
To find the baseball player's speed immediately after stopping the ball, we can use the principle of conservation of momentum. Since the ballplayer catches the ball while at the highest point of his leap, his vertical velocity is zero. The horizontal component of the ball's momentum is transferred to the ballplayer, resulting in his horizontal speed. The equation for conservation of momentum is:
m1v1 + m2v2 = m1v3 + m2v4
Where m1 and m2 are the masses (in kg) of the ballplayer and the ball respectively, and v1, v2, v3, and v4 are the initial horizontal velocity of the ball, final horizontal velocity of the ball, initial horizontal velocity of the ballplayer, and final horizontal velocity of the ballplayer respectively.
In this case, m1 = 79 kg, m2 = 0.14 kg, v1 = 25 m/s, and v3 = 0 m/s. By substituting these values into the equation, we can solve for v4:
(79 kg)(0 m/s) + (0.14 kg)(25 m/s) = (79 kg + 0.14 kg)(v4)
0 + 3.5 = 79.14 kg v4
v4 = 0.044 m/s
So the baseball player's speed immediately after stopping the ball is approximately 0.044 m/s.