Two particles are projected in air with speed `v_(0)` at angles `theta_(1)` and `theta_(2)` (both acute) to the horizontal,respectively.If the height reached by the first particle greater than that of the second,then thick the right choices

Options

A. angle of projection : `theta_(1) gt theta_(2)`
B. time of flight : `T_(1) gt T_(2)`
C. horizontal range : `R_(1) gt R_(2)`
D. total energy : `U_(1) gt U_(2)`.

Answer

A, B

Explanation

(a,b): Maximum height,
`h=(v_(0)^(2) sin ^(2) theta)/(2g)`
`therefore (h_(1))/(h_(2))=(sin^(2) theta_(1))/(sin^(2) theta_(2)) gt 1`
or `sin^(2) theta_(1) gt sin^(2) theta_(2)`
or `theta_(1) gt theta_(2)`
Also, time of flight , `T=(2v_(0)sin theta)/(g)`
`therefore (T_(1))/(T_(2))=(sin theta_(1))/(sin theta_(2)) gt 1` or `T_(1) gt T_(2)`
Similarly, horizontal range , `R=(v_(0)^(2) sin 2theta)/(g)`
`(R_(1))/(R_(2))=(sin 2 theta_(1))/(sin 2theta_(2)) gt 1` for `theta_(1), theta_(2) le (pi)/(4)`
`(R_(1))/(R_(2))=(sin 2 theta_(1))/(sin 2 theta_(2)) lt 1` for `theta_(1), theta_(2) gt (pi)/(4)`
The correct options are (a) and (b)