PhysicsThe equation of trajectory of a projectile thrown from a point on the ground is y=(x−x2/40)m. If g=10ms−2, the maximum height reached is

The equation of trajectory of a projectile thrown from a point on the ground is y=(xx2/40)m. If g=10ms2, the maximum height reached is

  1. A

    6 m

  2. B

    8 m

  3. C

    10 m

  4. D

    12 m

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    Solution:

    y=(xx240)    ;  g=10m/s2

    A=tanθ=1  

    θ=45º

    gu2cos2θ=140          

    u2=400, u=20m/s

    Maximum height (H)

    =u2sin2θ2g=20×202×10×(12)2=10m

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