Expert: Richard J. Raridon - 1/18/2007

Question
STONE OF MASS 0.62 FIRED UPWARDS, ANGLE TO GROUND, SPEED 5.5M/S.
KE AT MAXIMUM HEIGHT, 2.3J. HOW DO YOU WORK OUT THE HEIGHT ABOVE THE GROUND WHEN AT MAXIMUM POINT. THINK IVE GOT THE FIRST PART BUT UNSURE OF THE WHOLE EQUATION. ANY HELP WOULD BE MUCH APPRECIATED, THANKS

Answer
Since KE = (1/2)mv^2, v at max. height = 2.72m/s.  Since the vertical velocity at max. height = 0, the horizontal velocity throughout the flight has to be 2.72m/s.  Since the initial velocity = 5.5m/s, the initial vertical velocity = [5.5^2 -2.72^2]^1/2 = 4.78m/s.  
So 4.78^2 = 2gh and h = 1.17m