CATAGORY: Mechanics QUESTION: A ball is thrown straight upward and rises to a maximum height of 16 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half its initial value?

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We use the velocity vs. position (height) formula for a uniformly acceleratd object. (This is just the energy equation in disguise!) (1) Vf^2 -Vi^2 = -2gh Where Vf and Vi are the initial and final velocities. h is the difference in height and g = 9.8m/s^2. At h = 16m Vf = 0 m/s so (2) Vi^2 = 2gh or _____ (3) Vi = \/ 2gh = 17.7 m/s We now apply (1) again but this time let Vf = 1/2 Vi so that h is the height at which V is one half the inital value. Then we have (4) (1/2Vi)^2 -Vi^2 = -2gh or (5) -3/4 Vi^2 = -2gh or solving for h (6) 3/8 Vi^2 -------- = h = 11.99 m ~ 12m g *************** * * * h ~ 12.0m * * * ***************