T h e e q u a t i o n o f m o t i o n o f a p r o j e c t i l e i s y = 12 x - 3 4 x 2 . T h e h o r i z o n t a l c o m p o n e n t o f v e l o c i t y i s 3 m / s . W h a t i s t h e r a n g e o f t h e p r o j e c t i l e ? ( A ) 18 m ( B ) 16 m ( C ) 12 m ( D ) 21 . 6 m Share with your friends Share 0 Deepak Kumar answered this Dear Student , Given eqn of projectile motion, y = 12 x - 3 x24. Comparing this eqn with standard eqn of projectile motion. y = x tanθ - g x22 u2 cos2θ.We get tanθ = 12 and g2 u2 cos2θ=34 or u2 cos2θ = 10×42×3. or u cosθ = 2 23 . tanθ = sinθcosθ = 12 or u sinθ =u cosθ×12 or u sinθ = 24 23.Now the range of projectile = u cosθ× 2u sinθ/g = 2 23 × 2× 24 23 /10 = 64/3=21.33 m Regards 0 View Full Answer Siva Sekhar B R answered this Please find this answer 0 Juvi answered this Put y=0 and then solve it 0