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Question: Answered & Verified by Expert
A projectile is given an initial velocity of i^+3 j^ m s-1, where i^ is along the ground and j^ is along the vertical. Then, the equation of the path of projectile is [Take, g=10 m s-2
PhysicsMotion In Two DimensionsBITSATBITSAT 2019
Options:
  • A y=3x-5x2
  • B y=3x+5x2
  • C x=3y+5x2
  • D x2=y2+3
Solution:
1419 Upvotes Verified Answer
The correct answer is: y=3x-5x2

Velocity of projectile at time t is given by

v=u-g t j^

v=i^+3 j^-10 t j^

  r=v dt

= i^+3 j^-10 t j^ dt

=t i^+3 t j^-10t22j^

=t i^+3 t j^-5 t2 j^

r=t i^+3 t-5 t2 j^

or x i^+y j^=t i^+3 t-5 t2 j^

On comparing, x=t          ...(i)

and y=3 t-5 t2          ...(ii)

From Eqs (i) and (ii), we get

y=3 x-5 x2

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