Question

A 79 kg baseball player jumps straight up to catch a hard-hit ball.

If the 140 g ball is moving horizontally at 25 m/s , and the catch is made when the ballplayer is at the highest point of his leap, what is his speed immediately after stopping the ball?

Answer 1

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.

Answer 2

The speed of the player immediately after stopping the ball, given that the ball was initial moving at 25 m/s, is 0.044 m/s

How to calculate the speed immediately after stopping the ball?

The speed of the player immediately after stopping the ball is considered as the speed after collision. This can be calculated by using the conservation of linear momentum as illustrated below:

• Mass of baseball player (m₁) = 79 Kg
• Initial velocity of baseball player (u₁) = 0 m/s
• Mass of ball (m₂) = 140 = 140 / 1000 = 0.140 Kg
• Initial velocity of ball (u₂) = 25 m/s
• Speed after collision (v) = ?

`m_1u_1\ +\ m_2u_2 = (m_1\ +\ m_2)v\\\\(79\ *\ 0)\ +\ (0.140\ *\ 25) = (79\ +\ 0.140)v\\\\0\ +\ 3.5 = 79.14v\\\\3.5 = 79.14v\\\\v = (3.5)/(79.14) \\\\v = 0.044\ m/s`

Thus, we can conclude that the speed of player immediately after stopping the ball is 0.044 m/s