# Conservation of Energy (2-Dimensional!)

1. Oct 23, 2009

### interxavier

1. The problem statement, all variables and given/known data
A bullet is shot into the air with muzzle veloity Vo at an angle θ with the horizontal. Use energy considerations to find a) the highest point reached and b) the magnitude of the velocity when the bullet is at half its maximum height.

2. Relevant equations
Vx = Vo*cos(θ)
Vy = Vo*sin(θ)
@ Ymax, Vy = 0
Ei = Ef
KEi + PEi = KEf + PEf

3. The attempt at a solution
I got letter a by breaking Vo into its components.

1/2*m*Vo^2*cos^2(θ) + 1/2*m*Vo^2*sin^2(θ) + 0 = 1/2*m*Vo^2*cos^2(θ) + m*g*H
1/2*m*Vo^2*sin^2(θ) = mgH
H = Vo^2*sin^2(θ)/

My problem is with letter b. According to the textbook, the answer is:
V = sqrt(gh + Vo^2*cos^2(θ)) = sqrt(Vo^2/2[1 + cos^2(θ)]) =
Vo^2*sqrt(1 - 1/2*sin^2(θ))

I'm finding it difficult to getting to that answer. I basically used the same process as letter a except H is now H/2 and the final y-component of V is now in the final energy.

I ended up getting V = sqrt(Vo^2 - gH), which is obviously not consistent with the book's answer.

2. Oct 23, 2009

### pc2-brazil

Good evening,

In a, didn't you mean to write H = Vo^2*sin^2(θ)/2g?
The expression you obtained in a --the highest point-- is the maximum height.
In b, you obtained the formula by setting height = H/2; the problem asks for the velocity at half the maximum height.
So, in the formula you got (V = sqrt(Vo^2 - gH)) try to replace H with the expression for the maximum height you found in a, and see if it matches the book's answer.