A particle is projected at an angle theta with the horizontal . If angle of elevation of highest point of trajectory is phi when seen from point of projection, then

A particle is projected at an angle theta with the horizontal . If ang

1.	A particle is projected at an angle theta with the horizontal . If angle of elevation of highest point of trajectory is phi when seen from point of projection, then
Answer» `tan PHI =2 tan phi` `tan theta =3 tan phi` `tan theta =2 tan phi` `tan theta=4 tan phi` Solution :(c): Let particle be projected with a SPEED u From the FIGURE, `tan phi =(H)/(R//2)=(2H)/(R)=((2(u^(2) sin^(2)theta)/(2g))/(u^(2) sin 2 theta))/(g)` `tan phi =(sin^(2) theta)/(sin 2THETA)=(sin ^(2)theta)/(2 sin theta COS theta)=(1)/(2)tan theta` Hence, option (c) is correct .