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Step-by-Step Solution
Step 1: Identify the Forces Acting on the Ball
When the ball is thrown vertically upward with initial velocity $u$, it experiences two downward forces:
Gravitational force $mg$.
Resistive force $mk v^2$, where $v$ is the speed of the ball.
Taking the upward direction as positive, the net force $F_{\text{net}}$ acting on the ball at any instant is
$$
F_{\text{net}} = -\,mg \;-\; m k v^2.
$$
Step 2: Write Down the Acceleration Equation
Using Newton's second law ($F = m a$), the acceleration $a = \frac{dv}{dt}$ is:
$$
m\,\frac{dv}{dt} \;=\; -\,mg \;-\; m\,k\,v^2
\quad\Longrightarrow\quad
\frac{dv}{dt} \;=\; -\,g \;-\; k\,v^2.
$$
Step 3: Relate Velocity and Position
To find the displacement (height) as a function of velocity, use
$$
\frac{dv}{dx} \;=\; \frac{\frac{dv}{dt}}{\frac{dx}{dt}} \;=\; \frac{\frac{dv}{dt}}{v}.
$$
Since $\frac{dv}{dt} = -\,g - k\,v^2$, we get
$$
\frac{dv}{dx}
\;=\; \frac{-\,g - k\,v^2}{v}.
$$
Rearrange to separate the variables:
$$
\frac{v\,dv}{g + k\,v^2} \;=\; -\,dx.
$$
Step 4: Integrate to Find the Maximum Height
The ball’s velocity decreases from $v = u$ (initially) to $v = 0$ (at the maximum height $H$). The displacement changes from $x = 0$ to $x = H$. Therefore, integrate both sides:
$$
\int_{v=u}^{v=0} \frac{v\,dv}{g + k\,v^2}
\;=\;
- \int_{0}^{H} dx.
$$
The integral on the left is evaluated as:
$$
\int \frac{v\,dv}{g + k\,v^2}
\;=\; \frac{1}{2k}\;\ln\!\bigl(g + k\,v^2\bigr).
$$
Applying the limits $v = u$ to $v = 0$:
$$
\left[\frac{1}{2k}\,\ln\bigl(g + k\,v^2\bigr)\right]_{u}^{0}
\;=\;
\frac{1}{2k}\,\ln\!\bigl(g + k\,(0)^2\bigr)
\;-\;
\frac{1}{2k}\,\ln\!\bigl(g + k\,u^2\bigr).
$$
This simplifies to:
$$
\frac{1}{2k}\,\ln(g)
\;-\;
\frac{1}{2k}\,\ln\bigl(g + k\,u^2\bigr).
$$
Meanwhile, the right-hand side is
$$
-\int_{0}^{H} dx
\;=\; -\,H.
$$
Equating both sides:
$$
\frac{1}{2k}\,\ln(g)
\;-\;
\frac{1}{2k}\,\ln\bigl(g + k\,u^2\bigr)
\;=\; -\,H.
$$
Rewrite and factor out $-\frac{1}{2k}$:
$$
H
\;=\;
\frac{1}{2k}\,\ln\!\Bigl(\frac{g + k\,u^2}{g}\Bigr)
\;=\;
\frac{1}{2k}\,\ln\!\Bigl(
1 + \frac{k\,u^2}{g}
\Bigr).
$$
Step 5: State the Final Result
Therefore, the maximum height attained by the ball is
$$
H
\;=\;
\frac{1}{2k}\,\ln\!\Bigl(
1 + \frac{k\,u^2}{g}
\Bigr),
$$
which matches the given correct answer.
Reference Image
