# Conservation of energy in projectiles

1. Nov 28, 2004

### disruptors

an object launched vertically upward with an initial speed v. Neglect air resistance

a)At what height h above the ground does the projectile have a speed of 0.5v? Express your answer in terms of v and the magnitude of the acceleration of gravity g.

Using conservation of energy i found the max height (hmax) to be v^2/(2*g)

i then came to find h at a speed at 0.5v to be (v^2/(2*g))/4. However it says i am off by a multiplicative factor. I tried to take in account the KE and PE but am confused...

b) 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 thought it would be v^2/(2*g) and its not which makes me wonder why?

thanks

2. Nov 28, 2004

### marlon

I got :
$$\frac{mv^2}{2} = \frac{1}{2}*m* \frac{v^2}{2^2} + mgh$$

solve this for h...

at the max heigth, the y-component of the velocity (=v*sin(theta)) is 0.

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

regards

marlon