To practice problem-solving strategy 11.1 for conservation of momentum problems. 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?

0.081 m /s

Explanation:

According to the conservation of momentum, the momentum of a system is conserved when no external force is applied on a body.

momentum of the system before throwing = momentum of the system after throwing

Let v be the velocity of quarterback after throwing the football.

80 x 0 + 0.43 x 0 = 80 x v + 0.43 x 15

0 = 80 v + 6.45

v = - 0.081 m /s

The negative sign shows that he is moving in backward direction.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

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i just did this and got a 100.