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?

The speed of ballplayer during taking the catch is 0.049 m/s.

Given data:

The mass of ball is, m = 140 g = 0.140 kg.

The speed of ball is, 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.

Also, let a 65 kg baseball player jumps straight up to catch a hard-hit ball. Then according to the conservation of linear momentum,

Here, u' is the initial speed of player. Since, the player was initially at rest, then u' = 0.

Solving as,

Thus, we can conclude that the speed of ballplayer is 0.049 m/s.

Learn more about the conservation of momentum here:

link

0.049 m/s

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 :

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

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray