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

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`

Step-by-step 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) = (m_Bv_(Bx))/(m_Q)`

substituting values

`v_q = (0.43 * 15)/(80)`

`v_q = 0.08 \ m/s`