# Bullet hits the ground on its return

1. Feb 7, 2012

### SAHM1500

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:(

2. Feb 7, 2012

### Brunetto

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...

3. Feb 8, 2012

### SAHM1500

this is may be more clear.
thanks.

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4. Feb 8, 2012

### Brunetto

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...

5. Feb 8, 2012

### SAHM1500

Yes i did 2.12 .. I proved *the equations*!

6. Feb 8, 2012

### SAHM1500

Please can any body help me!!!