Conservation of Energy (2-Dimensional)

[interxavier]

Homework Statement

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.

Homework Equations

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

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.

 

[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.

 

[kerimek]

Can someone please help me understand where I went wrong and how to get the correct answer?


The approach you took for part a is correct, but for part b, you need to consider the conservation of energy at the half maximum height, not at the highest point. This means that the potential energy at half maximum height is equal to half of the potential energy at the highest point. Therefore, your equation should be:

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

Solving for V, we get:

V = sqrt(gh + Vo^2*cos^2(θ)) = sqrt(Vo^2/2[1 + cos^2(θ)]) = Vo*sqrt(1 + cos^2(θ)/2)

This is consistent with the answer given in the textbook. The key is to remember that at the half maximum height, the potential energy is half of the potential energy at the highest point, and this needs to be taken into account in the conservation of energy equation.