From the ground an object is projected upward with sufficient velocity so that it crosses the top of a tower in time t1 and reaches the maximum height. It then comes down and recrosses the top of the tower in time t2 , time being measured from the instant the object was projected up. A second object released from the top of the tower reaches the ground in time t3 . Show that t3 = sqrt(t1·t2) .Solution: |

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