SOLUTION: A juggler threw an orange upward while standing on a platform 2.00 m above the floor with a velocity of 2.00 m/s. Neglect air resistance. a. How high does the orange rise? b. H

Question 1180007: A juggler threw an orange upward while standing on a platform 2.00 m above the floor with a velocity of 2.00 m/s. Neglect air resistance.
a. How high does the orange rise?
b. How long does it take for the juggler to catch the orange?
c. How fast was the orange moving just before the juggler caught it?
Answer by ikleyn(52814) (Show Source): You can put this solution on YOUR website!
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(a)  t = v/g,  v is initial velocity (2 m/s);  g is the gravity acceleration (10 m/s^2).


     So, t = 2/10 = 0.2 seconds.


     " How high "   is  the product of the time of 0.2 seconds by the average speed of 2/2 = 1 m/s, i.e.  0.2*1 = 0.2 meters.



(b)  Two times this time: 0.2 seconds + 0.2 seconds = 0.4 seconds.



(c)  the speed of the orange at the caugh moment is the same as the initial speed, 2 m/s  (energy conservation law)

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