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_`
D. horizontal range : ` R_1 gt R_2`

Answer

A, B

Explanation

Height, ` h= ( v_0^2 sin^2 theta)/( 2g) , i.e., h prop sin^@ theta`
:. H_1 /h_2 = ( sin^2 theta_1)/( sin^@ theta_2 gt 1` so sin^2 theta_1 gt sin^2 theta_1 gt sin^2 theta_2 ` or theta_1 gt theta_2`
Time of fight, ` T= ( 2 v_0 sin theta)` or ` T prop sin theta)`
:. ` T_1 /T_2 = ( sin theta_1)/( sin theta_2) ` or T_1 gt T_2`
Horizontal range, ` R( u^2 sin 2 hteta)/g or ` R prop sin 2 theta`
:. ` R_1 /R_2 = ( sin 2 theta_1)/( sin 2 theta_2) lt 1` or R_1 -ltR_2`
Total energy of each particle will be equal to KE of each jparticle at the tiem of its projection.