Trajectory of two particles projected from origin with speed $v_1$ and $v_2$ and angles ${\theta }_1$ and ${\theta }_2$ with positive x- axis respectively as shown in the figure given that $g=-10m/s^2\left(j\right)$. Choose the correct option related to diagram :

The correct options are
B $3(v_1-v_2)=0$
D $3({\theta }_1-{\theta }_2)=-{\theta }_1$
Horizontal range $R=\frac{v^{2}sin\left(2\theta \right)}{g}=\frac{2v^{2}sin\theta cos\theta }{g}$

Maximum height attained $H=\frac{v^{2}si{n}^2\theta }{2g}$

For particle 1 : $R_1=40$ m $H_1=10$ m

$\therefore$ $10=\frac{v_1^{2}si{n}^2{\theta }_1}{2g}$ ..........(1)

And $40=\frac{2v_1^{2}sin{\theta }_{1}cos{\theta }_1}{g}$ .............(2)

Dividing (1) by (2), we get $1=tan{\theta }_1$ $⟹{\theta }_1={45}^o$

$\therefore$ From (1) we get $10=\frac{v_1^2\times 0.5}{2\left(10\right)}$ $⟹v_1=20m/s$

For particle 2 : $R_2=20\sqrt{3}$ m $H_2=15$ m

$\therefore$ $15=\frac{v_2^{2}si{n}^2{\theta }_2}{2g}$ ..........(3)

And $20\sqrt{3}=\frac{2v_2^{2}sin{\theta }_{2}cos{\theta }_2}{g}$ .............(4)

Dividing (3) by (4), we get $\sqrt{3}=tan{\theta }_2$ $⟹{\theta }_2={60}^o$

$\therefore$ From (3) we get $15=\frac{v_2^2\times (0.866{)}^2}{2\left(10\right)}$ $⟹v_2=20m/s$

$\therefore$ $3({\theta }_1-{\theta }_2)+{\theta }_1=3({45}^o-{60}^o)+{45}^o=0$