How Fast Will Zeke Slide Down the Hill?

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Zeke, weighing 76 kg with a 2 kg rubber mat, is sliding down a snow hill, and the problem asks for his speed 1.2 m below the crest, ignoring friction. The initial calculation used the formula v = √(2gh) but questioned the accuracy of the result, which was 4.85 m/s. A key point of confusion arises regarding whether the 1.2 m is measured vertically or along the slope of the hill. If it's vertical, the equation is mostly correct, but the negative sign under the square root is problematic. Clarification on the measurement is crucial for determining the correct approach to solving the problem.
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Homework Statement


Zeke begins to slide down a snow hill on a rubber mat. Zeke's mass is 76 kg and that of the mat is 2kg.



Homework Equations


Disregarding frictional forces, how fast are they moving when they are 1.2 m below the crest?



The Attempt at a Solution


v=(square root)2gh
v=(square root)(2)(9.8)(-1.2)
v=4.85 m/s

I don't think this is even close to being right, as a friend who I just got done talking to (and is 10x smarter than me) said he got a completely different answer.

By the way, hello everyone! I just signed up, hoping to learn to understand physics better.
 
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Hello, we do hope you understand better from now on.

Does the problem mention 1.2 m vertically below the crest or 1.2 m along the slope?

If it's the first case, your eqn is correct, except for the minus sign under the sqrt. Is that possible?

If it's the second case, you'll have to take the component of g along the slope.
 
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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