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why the path of projectile is parabola and not as like horizontal?
@ gopala suppose a projectile of mass m is projected with velocity u at an angle theta with the ground . the horizontal component = u cos theta ( remain constant ) the horizontal displacement at any time t is x = uxt = ut cos theta t = x/ u cos theta the vertical component is given by u (y ) = u sin theta the vertical displacement at any time t is Y = XTAN THETA – G X SQUARE / { 2 U SQUARE COS SQUARE THETA } and , this above equation is a equation of a parabola . so, the path of projectile is a parabola HOPE IT CLEARS ALL THE BEST ..
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