Kinematics

JEE Main level, single correct, medium

130 of 308

Question 130 of 308

A small ball of mass mm is thrown upward with velocity uu from the ground. The ball experiences a resistive force mkv2mk{v}^{2} where vv is it speed. The maximum height attained by the ball is :

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Answer: (D) 12kln⁡(1+ku2g)\dfrac{1}{2k}\ln \left(1+\dfrac{k{u}^{2}}{g}\right)

Fne=ma{F}_{\text{ne}}=ma
−mg−mkv2=mvdvds-mg-{mkv}^{2}=mv\dfrac{dv}{ds}
vdvds=−g−kv2v\dfrac{dv}{ds}=-g-{kv}^{2}
−∫v00vdvg+kv2=∫0hds=hmax⁡-{\int }_{{v}_{0}}^{0}\dfrac{vdv}{g+{kv}^{2}}={\int }_{0}^{h}ds={h}_{\max }
hmax⁡=12Kin(g+kv02g){h}_{\max }=\dfrac{1}{2K}in\left(\dfrac{g+{kv}_{0}^{2}}{g}\right)

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