Question

SOMEONE HELP  ASAP PLSS

Answer 1

–0.05 m/s

Explanation:

The total momentum of the system player+basketball must be conserved before and after the ball has been thrown.

Before throwing the ball, the total momentum of the system is zero, because can assume both the player and the basketball being at rest:

$p_i$

The total momentum after the ball has been thrown is instead the sum of the momenta of the the player and of the basketball:

$p_f=m_p v_p + m_b v_b$

where

$m_p = 59 kg$ is the player's mass

$v_p$ is the player's velocity

$m_b=0.51 kg$ is the ball's mass

$v_b=6 m/s$ is the ball's velocity

For the conservation of momentum, we have

$p_i=p_f$

$0=m_p v_p + m_b v_b$

$v_p=-\frac{m_b v_b}{m_p}=-\frac{(0.51 kg)(6 m/s)}{59 kg}=-0.05 m/s$

And the negative sign means that the player travels in the opposite direction to the ball.

Answer 2

We apply the momentum preservation principle and we have p before =p after
0=6*0.51+59u
u=-0.05m/s