Answer to Question #23335 in Mechanics | Relativity for Gabriel Marin
A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant a ball is dropped from rest from a building 14 m high. After how long will the balls be at the same height?
Сoordinate of the first ball depends on time as follows: x1 = 19.2t - 9.8t^2/2 19.2 - initial speed 9.8 - gravitational acceleration Сoordinate of the second ball depends on time as follows: 14 - 9.8t^2/2 14 - initial coordinate When the balls will be at the same height their coordinates will be equal: x1 = x2: 19.2t - 9.8t^2/2 = 14 - 9.8t^2/2 t = 14/19.2 = 0.73 seconds Answer: 0.73 sec
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