Work and acceleration of a touchdown

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The discussion revolves around calculating the work done by a wide receiver who leaps 1.20 m after scoring a touchdown. The work is calculated using the formula W = FD, resulting in 988 J when applying the receiver's mass and gravitational acceleration. There is confusion regarding the acceleration value of 9.8 m/s², with questions about why it is considered positive despite gravity acting downwards. It is clarified that while the receiver's upward motion is opposed by gravity, the acceleration due to gravity remains a constant downward force. Ultimately, the downward acceleration acts as a decelerating force during the jump, affecting the receiver's upward velocity.
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Homework Statement


After scoring a touchdown, an
84.0-kg wide receiver celebrates by leaping
1.20 m off the ground. How much work
was done by the wide receiver in the
celebration?


Homework Equations


W= FD


The Attempt at a Solution


I know that the correct answer should be:
(84.0 kg)(9.80 m/s2)(1.20 m)=988 J
I'm just wondering why the acceleration for the wide-receiver is 9.8m/s^2.
If his acceleration is 9.8 and the acceleration of gravity is 9.8, wouldn't the total acceleration be 0? Also, when a person jumps, he starts off with a positive velocity which decreases, so shouldn't his acceleration be negative?
 
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Once he is in the air, the only force acting on him is his weight i.e. the pull of gravity on him. But weight W = mg.

Hence if we use F = ma (downwards) for the man

we get

W = ma
mg = ma
i.e.g=a.
 
OK I get that the acceleration should be 9.8, because there's only one force, but why is it positive instead of negative?
 
Last edited:
The acceleration due to gravity is always towards the centre of the Earth. In our case we can say that the acc due to gravity is always downwards.

Hence if an object is going upwards this DOWNWARD acceleration due to gravity will decrease this velocity and so it acts as a retardation while if the object is going downwards this DOWNWARDS acceleration due to gravity will increase the velocity.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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