Question

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 mγν²
(where m is mass of the ball, v is its Instantneous velocity and γ is a constant). Time taken by the ball to rise to its zenith is:
(A) {1/√(γg)} ln{1+ √(γ/g) V₀}
(B) {1/√(γg)} tan⁻¹{√(γ/g) V₀}
(C) {1/√(γg)} sin⁻¹{√(γ/g) V₀}
(D) {1/√(γg)} tan⁻¹{√(γ/g) V₀}

Answer 1

Explanation:

cant understand that (A) (B) (C) (D)

Answer 2

Time taken by the ball to rise to its zenith is t = $\sqrt{\frac{1}{gY} } tan^{-1} \sqrt{\frac{Y}{g} }V0$

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 mγν²

.

The net force ma on the body is,

ma = -mg  - mγν²

==> a = -(g + γν²) = $\frac{dv}{dt}$

==> -gdt = dv/(1 + γν²/g) --(1)

integrating both sides t : 0 --> t and V: V0 --> 0,

$\int\limits^x_0 {\frac{1}{1+x^{2} } } \, dx =$ $tan^{-1}$x

by substitution, and integrating (1),

-gt = - $\sqrt{\frac{g}{Y} } tan^{-1} \sqrt{\frac{Y}{g} }V0$

==> t = $\sqrt{\frac{1}{gY} } tan^{-1} \sqrt{\frac{Y}{g} }V0$