Question

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?

Answer 1

The speed of the ballplayer after catching the ball is 25 m/s.

Step-by-step explanation:

The speed of the ballplayer after catching the ball can be found using the principle of conservation of momentum. The momentum before the catch is equal to the momentum after the catch. The momentum of the ball before the catch is calculated by multiplying its mass (140 g) by its velocity (25 m/s). Since the ballplayer is at the highest point of his leap when he catches the ball, his vertical velocity is zero. Therefore, the speed of the ballplayer after catching the ball is 25 m/s.

Answer 2

0.049 m/s

Step-by-step explanation:

Let a 71 kg baseball player jumps straight up to catch a hard-hit ball.

Mass of the player, m = 71 kg

Mass of the ball, m' = 140 g = 0.14 kg

Initial velocity of the player, u = 0 (at rest)

Initial velocity the ball, u' = 25 m/s

We need to find the speed when the ballplayer is at the highest point of his leap. They will stick with each other. Let they move with the speed of V.

Using the conservation of momentum as follows :

`mu+mu'=(m+m')V\\\\V=(mu)/((m+m'))\\\\V=(0.14* 25)/((71+0.14))\\\\V=0.049\ m/s`

So, his speed at the highest point is 0.049 m/s.