show that the path of projectile is parabola when it is projected from light
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Suppose a projectile of mass ‘m’ is projected with velocity ‘u’ at an angle θ with the ground.
The horizontal component of the initial velocity is, ux = u cosθ [this remains constant]
The horizontal displacement at any time ‘t’ is, X = ux t = ut cosθ
=> t = X/(u cosθ)
The vertical component is given by, uy = u sinθ
The vertical displacement at any time ‘t’ is, Y = uy t – ½ gt 2 = ut sinθ – ½ gt2
=> Y = u[X/(u cosθ)]sinθ – ½ g[X/(u cosθ)]2
=> Y = X tanθ – gX2 /[2u2 cos2 θ]
Which is the equation of a parabola. So, the path of the projectile is parabolic.
The horizontal component of the initial velocity is, ux = u cosθ [this remains constant]
The horizontal displacement at any time ‘t’ is, X = ux t = ut cosθ
=> t = X/(u cosθ)
The vertical component is given by, uy = u sinθ
The vertical displacement at any time ‘t’ is, Y = uy t – ½ gt 2 = ut sinθ – ½ gt2
=> Y = u[X/(u cosθ)]sinθ – ½ g[X/(u cosθ)]2
=> Y = X tanθ – gX2 /[2u2 cos2 θ]
Which is the equation of a parabola. So, the path of the projectile is parabolic.
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