Question

An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?PLEASE SORT QUESTION BELOWSort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.mQ: the mass of the quarterbackmB: the mass of the football(vQx)i: the horizontal velocity of quarterback before throwing the ball(vBx)i: the horizontal velocity of football before being thrown(vQx)f: the horizontal velocity of quarterback after throwing the ball(vBx)f: the horizontal velocity of football after being thrown

Answer 1

Answer:

a

The speed of the quarterback backward is $v_q = 0.08 \ m/s$

b

Known are

 $m_Q , m_B , (v_{Bx})_i (v_{Qx})_f, (v_{Bx})_f$

Unknown

   $(v_{Qx})_f$

Explanation:

From the question we are told that

   The mass of the quarterback is $m_Q = 80 \ kg$

    The mass of the ball is $m_B = 0.43 \ kg$

     The speed of the ball is  $v_{B x}= 15 \ m/s$

The law of momentum conservation can be mathematically represented as

       $m_Q u_{Qx} + m_Bu_{Bx} = - m_{Q} v_{Qx} + m_B v_{Bx}$

Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball

  So

       $m_Q v_{Qx} = m_B v_ {Bx}$

=>     $v_{Qx} = \frac{m_Bv_{Bx}}{m_Q}$

substituting values

        $v_q = \frac{0.43 * 15}{80}$

       $v_q = 0.08 \ m/s$