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
             
             
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         *             *
         *  h ~ 12.0m  *
         *             *
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