InterviewSolution
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InterviewSolution
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When a particle is projected from the surface of earth its mechanical energy and angular momentum about center of earth at all time are constant. Q. Maximum height `h` of the particle isA. `=(v_(0)^(2)sin^(2)theta)/(2g)`B. `gt(v_(0)^(2)sin^(2)theta)/(2g)`C. `lt(v_(0)^(2)sin^(2)theta)/(2g)`D. can be greater than or less than `(v_(0)^(2)sin^(2)theta)/(2g)` |
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Answer» Correct Answer - B By applying conservation energy `(1)/(2)mv_(0)^(2)-(GM_(e)m)/(r)=(1)/(2)mv^(2)-(GM_(e)m)/((R+h))` Solving above equation `hgt(v_(0)^(2)sin^(2)theta)/(2g)` Alternate: As height increases gravitational force decreases and hence the acceleration therefore height will be more than `H=(v_(0)^(2)sin^(2)theta)/(2g)` |
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