Bullet hits the ground on its return

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The discussion focuses on deriving the speed of a bullet upon its return to the ground, using the equations for upward and downward motion. It highlights that the speed at impact can be expressed as Vo Vt / (Vo^2 + Vt^2)^0.5, where Vo is the initial speed and Vt is the terminal speed. The conversation reveals frustration over the complexity of the problem and a request for clearer formatting to facilitate understanding. Participants are also seeking assistance with a related question (2.12) that references the derived equations. Overall, the thread emphasizes the challenge of solving the problem and the need for collaborative support.
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V^2=Ae^-2kx - g/k (upward motion)
V^2=g/k - Be^2kx (downward motion)

Use the above result to show that , when the bullet hits the ground on its return, the speed will be equal to the expression

Vo Vt / (Vo^2 + Vt^2)^0.5

In which Vo is the initial upward speed and Vt= (mg/C2)^0.5= terminal speed = (g/k)^0.5

This is result allows one to find the fraction of the initial kinetic energy lost through air friction

THIS PROBLEM WILL KILL ME I SPEND DAYS TO SOLVE IT BUT I COULD NOT:(
 
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Can you put this question in a more formatted manner? It's rather hard to follow which is probably why people are viewing but not repsonding...
 
this is may be more clear.
thanks.
 

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did you solve question 2.12? if so what was your answer? 2.12 didn't come out clear enough for me to solve on my own. The question says to use that answer (not the equations) to solve it...
 
Yes i did 2.12 .. I proved *the equations*!
 
Please can anybody help me!
 
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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