The equation of trajectory of a projectile fired horizontally from a tower with Velocity U is given by:
(Take vertically upward as + Y- axis and horizontal as X-axis )

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$y = \frac{g}{2 U^{2}} x^{2}$

$y = \frac{- g}{2 U^{2}} x^{2}$

$x = \frac{g}{2 U^{z}} y^{2}$

$x = \frac{- g}{2 U^{z}} y^{2}$

answer is B.

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Detailed Solution

$l e t h o r i z o n t a l c o m p o n e n t o f v e l o c i t y o f b o d y p r o j e c t e d h o r i z o n t a l l y b e u = u_{x} v e r t i c a l c o m p o n e n t o f v e l o c i t y o f b o d y = u_{y} = 0 a f t e r t i m e t s e c o n d, d i s p l a c e m e n t o f t h e b o d y a l o n g h o r i z o n t a l i s x = u_{x} t \Rightarrow t = \frac{x}{u_{x}} - - - (1) d i s p l a c e m e n t o f t h e b o d y a l o n g v e r t i c a l i s y = u_{y} t + \frac{1}{2} (- g t^{2}) = 0 + \frac{1}{2} (- g t^{2}) - - - (2) s u b s t i t u t e e q u a t i o n (1) i n (2) y = \frac{- 1}{2} g {(\frac{x}{u_{x}})}^{2} = \frac{- 1}{2} g {(\frac{x}{u})}^{2} t h i s e q u a t i o n i s i n t h e f o r m o f y = k x^{2} w h i c h r e p r e s e n t s e q u a t i o n o f a p a r a b o l a$