A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a resistive force mkv² where v is it speed. The maximum height attained by the ball is:

Option 1 -

1/g ln(1 + ku²/2g)

Option 2 -

1/2k tan⁻¹(ku²/g)

Option 3 -

1/2k tan⁻¹(ku²/2g)

Option 4 -

1/2k ln(1 + ku²/g)

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

Mg + MkV² = MA = -mv (dV/dx)
Vdv = (−) (g + kV²)dx
∫? (Vdv)/ (g + kv²) = ∫? - dx
[ln (g + kV²)/2k]? = -x
ln (g/ (g + ku²) = −2kx
x = (1/2k)ln (1 + ku²/g)

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