Answer:
vBxf = 0.08625m/s
Explanation:
This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

pf: final momentum
pi: initial momentum
The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.
Then, you have:
(1)
mQ: the mass of the quarterback = 80kg
mB: the mass of the football = 0.43kg
(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s
(vBx)i: the horizontal velocity of football before being thrown = 0m/s
(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?
(vBx)f: the horizontal velocity of football after being thrown = 15 m/s
You replace the values of the variables in the equation (1), and you solve for (vBx)f:

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.
Hence, the speed of the quarterback after he throws the ball is 0.08625m/s
Use the definition of average acceleration:
<em>a</em> = ∆<em>v</em> / ∆<em>t</em>
We have
<em>a</em> = (8.29 m/s - 3.29 m/s) / (4.99 s)
<em>a</em> = 1.002 m/s²
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