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Question

A projectile is given an initial velocity of  ^i+2^j. The Cartesian equation of its path is 
(Take g=10 m/s2)
  1. y=2x5x2
  2. y=x5x2
  3. 4y=25x2
  4. y=2x25x2


Solution

The correct option is A y=2x5x2
Initial velocity of projectile =^i+2^jux=1,uy=2
We know that
Range of a projectile, R=u2sin2θg=2(ucosθ)(usinθ)gR=2uxuyg=2×1×210=25tanθ=uyux=21=2
Using equation of trajectory of a projectile, 
y=xtanθ(1xR)y=2x(15x2)=2x5x2
Alternate Solution:
ux=1,uy=2tanθ=uyux=2u=u2x+u2y==5
Equation of trajectory of a projectile
y=xtanθgx22u2cos2θ
=2x10x22×5×(15)2
i.e. y=2x5x2

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