An object of mass m is thrown vertically up. In the presence of heavy air resistance the time of ascent (t1) is no longer equal to the time of descent (t2). Similarly the initial speed (u) with which the body is thrown is not equal to the final speed (v) with which the object returns. Assuming that the air resistance F is constant show that t2/t1=sqrt((g+F/m)/(g-F/m)); v/u=sqrt((g+F/m)/(g-F/m));Solution: |

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