Question

A 72 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 140 g ball, which was moving horizontally at 25 m/s, while at the highest point of his leap is approximately 0.0486 m/s. This is calculated using the conservation of momentum.

Step-by-step explanation:

The problem involves the principle of conservation of momentum because no external horizontal forces are acting on the system of the baseball player and the ball during the catch. We first calculate the initial momentum of the ball and then set it equal to the final combined momentum of the ball and the baseball player to find the player's velocity after stopping the ball.

The initial horizontal momentum of the ball (pball_initial) is given by:

mball × vball_initial where mball is the mass of the ball and vball_initial is the initial horizontal velocity of the ball.

The final momentum of the combined system (player + ball) after the catch (pfinal) is given by:

(mplayer + mball) × vplayer_final where mplayer is the mass of the player, mball is the mass of the ball, and vplayer_final is the horizontal velocity of the player after making the catch.

Using conservation of momentum:

mball × vball_initial = (mplayer + mball) × vplayer_final

Solving for vplayer_final, we get:

vplayer_final = × mball × vball_initial / (mplayer + mball)

Plugging in the given values:

vplayer_final = (0.140 kg × 25 m/s) / (72 kg + 0.140 kg)

vplayer_final is approximately:

0.0486 m/s