A projectile is thrown at angle θ and 900–θ from the same point with same velocity 98 m/s. If the difference of heights is 50 m, the heights attained by them  will be- (in m)

maximum height reached by projectile is h1=u2sin2θ2g and h2=u2sin2(90−θ)2g∴h1+h2=u22gsin2θ+cos2θu22g=9822×10=490=h1+h2; given u=98m/s   and        h1−h2=50 ; solving these equations ∴h1=270m and h2=220m