# Projectile and Conservation of Energy

1. Nov 28, 2004

### Heart

Hi,

I need help; here's the question:

"A ball is launched as a projectile with initial speed v at an angle theta above the horizontal. Using conservation of energy, find the maximum height (hmax) of the ball's flight. Express your answer in terms of v, g, and theta."

I tried hmax = ((1/2)*(v^2))/(g) which didn't work; I then realized that it's not a straight throw the ball upward question.
I tried hmax = ((1/2)*((v)*sin(theta)))^2))/(g) and it didn't worked either; I thought I might have mistyped so I answered that question again, with the same answer.
I then got very frustrated and impulsive since I had only 2 more trys, so I tried hmax = ((1/2)*(vcos(theta)^2))/(g), which is obviously wrong and didn't work.

2. Nov 28, 2004

### marlon

at max heigth, the y-component of the velocity is ZERO but the x component is $$v_x = v*cos(\theta)$$.

Via conservation of energy we can calculate the maximal height. We know v which is the initial velocity and i suppose you can say that the object is launched from the x-axis so the initial height is zero. (if this is not the case, the an initial value must have been given, nevertheless the algorithm remains the same.) Now let's evaluatue the total energy at two different heights : the maximal height and the initial height. The conservation means that the total energy is the SAME, so we have that : $$mgh_{max}+\frac{m{v_x}^2}{2} = mg*0 + \frac{m{v_{initial}}^2}{2}$$ Now solve this for h and the problem is solved...

regards

marlon

3. Nov 28, 2004

### Heart

But wouldn't vinitial also = vcos(theta) = to vmax as well, in the x-axis, since this is the projectile motion? Shouldn't vinitial and vmax in the x-axis cancel each other out? and that only leaves us with vinitial in the y-axis, while vmax in the y-axis = 0? (and vinitial in the y-axis = v*sin(theta))

4. Nov 28, 2004

### marlon

No absolutely not...

at maximal height the y-component (perpendicular to the horizontal x-axis) needs to be zero because you are no longer rising. The x-component of the velocity remains constant during the entire motion because gravity only works in the y-direction so that all forces in the x-direction are 0...

marlon

5. Nov 28, 2004

### marlon

$$mgh_{max}+\frac{m{v_x}^2}{2} = \frac{m{v_{initial}}^2}{2}$$

and

$${v_{initial}}^2 = v^2$$

so that

$$mgh_{max} = \frac{m}{2} * v^2 (1 - cos^2(\theta))$$

regards
marlon