Answer:
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 :
[tex]mu+mu'=(m+m')V\\\\V=\dfrac{mu}{(m+m')}\\\\V=\dfrac{0.14\times 25}{(71+0.14)}\\\\V=0.049\ m/s[/tex]
So, his speed at the highest point is 0.049 m/s.
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,
[tex]mu+Mu'=(m+M)V[/tex]
Here, u' is the initial speed of player. Since, the player was initially at rest, then u' = 0.
Solving as,
[tex](0.140 \times 25)+(65 \times 0)=(0.140+65)V\\\\\V= \dfrac{(0.140 \times 25)}{(0.140+65)}\\\\V = 0.049 \;\rm m/s[/tex]
Thus, we can conclude that the speed of ballplayer is 0.049 m/s.
Learn more about the conservation of momentum here:
https://brainly.com/question/18066930
