Final answer:
The quarterback will move backward at approximately 0.081 m/s after throwing the football due to the conservation of momentum, which indicates that the total momentum before and after an event in a closed system stays constant if no external forces act on it.
Step-by-step explanation:
The question concerns a physics concept known as the conservation of momentum. In the scenario, an 80-kg quarterback jumps straight up and throws a 0.43-kg football horizontally at 15 m/s. To find out how fast he will be moving backward just after releasing the ball, we can use the conservation of momentum principle, which states that the total momentum of a closed system before an event is equal to the total momentum after the event, provided no external forces are acting on it. Since both the quarterback and the football are part of the system and initially the system's horizontal momentum is zero, the momentum gained by the ball must be equal in magnitude and opposite in direction to the momentum gained by the quarterback.
Using the formula for momentum (p = m * v), we have:
momentum of the football = mass of the football * velocity of the football = 0.43 kg * 15 m/s = 6.45 kg·m/s
Since momentum is conserved,
momentum of the quarterback = - momentum of the football
Therefore, the velocity of the quarterback (v_qb) can be found by rearranging the formula for momentum:
v_qb = - momentum of the football / mass of the quarterback = -6.45 kg·m/s / 80 kg
v_qb = -0.080625 m/s
This means that the quarterback will be moving backward at a velocity of approximately 0.081 m/s (negative indicating the opposite direction to the throw) just after releasing the ball.