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A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is affected by a drag force equal to mγv² (where m is mass of the ball, v is instantaneous velocity and γ is constant). Time taken by the ball to rise to its zenith is:

1. 1/√(γg) tan⁻¹(√(γ/g) V0)
2. 1/√(2γg) tan⁻¹(√(2γ/g) V0)
3. 1/√(γg) sin⁻¹(√(γ/g) V0)
4. 1/√(γg) ln(1 + √(γ/g) V0)

Correct answer: 1/√(γg) tan⁻¹(√(γ/g) V0)

Solution

Going up: m dv/dt = -mg - m*gamma*v^2. Then t = integral_0^{V0} dv/(g+gamma v^2) = (1/sqrt(gamma g)) * arctan(sqrt(gamma/g)*V0).

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