Respuesta :
Answer:
The player's speed afterward ball is thrown at 18 [tex]\frac{m}{s}[/tex] relative to ground is 2.04 [tex]\frac{m}{s}[/tex]
Explanation:
Given:
Mass of football player [tex]m = 71[/tex] kg
Speed of player [tex]v = 2.15[/tex] [tex]\frac{m}{s}[/tex]
Mass of football [tex]m' = 0.470[/tex] kg
For finding the player's speed afterward the ball is thrown at 18 [tex]\frac{m}{s}[/tex] relative to the ground,
The initial momentum of football player and ball is,
[tex]P_{i} = (m+m') v[/tex]
[tex]P_{i} = (71 +0.470) 2.15[/tex]
[tex]P_{i} = 153.66[/tex] [tex]kg . \frac{m}{s}[/tex]
The final momentum of the system is,
[tex]P_{f} = mv_{o} + m'v_{b}[/tex]
Here given in question [tex]v_{b} = 18 \frac{m}{s}[/tex]
[tex]P_{f} = 71 v_{o} + 0.470 \times 18[/tex]
Using the conservation of momentum,
[tex]P_{i} = P_{f}[/tex]
[tex]153.66 = 71 v_{o} + 0.470 \times 18[/tex]
[tex]71v_{o} = 145.2[/tex]
[tex]v_{o} = 2.04[/tex] [tex]\frac{m}{s}[/tex]
Therefore, the player's speed afterward ball is thrown at 18 [tex]\frac{m}{s}[/tex] relative to ground is 2.04 [tex]\frac{m}{s}[/tex]
