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 its speed. The maximum height attained by the ball is :

1 \over 2k\ln ( 1 + ku^2 \over g )

1 \over k\tan ^ - 1ku^2 \over 2g

1 \over 2k\tan ^ - 1ku^2 \over g

1 \over k\ln ( 1 + ku^2 \over 2g )

Laws of Motion PYQ — JEE Main 2020 · 4 Sep · Shift 2 · Q58 with Solution

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Correct answer: C

When the ball moves upward, both forces act downward:

So, taking upward as positive, $m\frac{dv}{dt}=-mg-mkv^2$

Dividing by $m$ , $\frac{dv}{dt}=-(g+kv^2)$

We need maximum height, so use $\frac{dv}{dt}=v\frac{dv}{dx}$

Thus, $v\frac{dv}{dx}=-(g+kv^2)$

Rearranging, $dx=-\frac{v\,dv}{g+kv^2}$

At launch:

At maximum height:

So,

=\int_0^u \frac{v\,dv}{g+kv^2}$$ --- 4. **Evaluate the integral** $$H=\int_0^u \frac{v\,dv}{g+kv^2}$$ Let $$t=g+kv^2 \Rightarrow dt=2kv\,dv$$ So, $$v\,dv=\frac{dt}{2k}$$ Hence, $$H=\frac{1}{2k}\int_{g}^{g+ku^2}\frac{dt}{t}$$ $$H=\frac{1}{2k}\left[\ln t\right]_{g}^{g+ku^2}$$ $$H=\frac{1}{2k}\ln\left(\frac{g+ku^2}{g}\right)$$ $$H=\frac{1}{2k}\ln\left(1+\frac{ku^2}{g}\right)$$ --- 5. **Match with the options** This is exactly: $$\boxed{\frac{1}{2k}\ln\left(1+\frac{ku^2}{g}\right)}$$ So the correct option is **C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They match.