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A particle of mass ' m ' is projected with a velocity v = kVe(k < 1) from the surface of the earth.
(Ve = escape velocity)
The maximum height above the surface reached by the particle is:

Option 1 -

R(k²/(1-k²))

Option 2 -

R((1-k²)/k²)

Option 3 -

R²k/(1+k)

Option 4 -

Rk²/(1-k²)

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 4


    Detailed Solution:

    h = R / [ (2gR/V? ²) - 1] = R / [ (V²/K²V? ²) ] = KK² / (1 - K²)

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