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 is

Options

A. `=(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)`

Answer

B

Explanation

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)`