Question

A small ball of mass $m$ is thrown upward with velocity $u$ from the ground. The ball experiences a resistive force $mkv^2$ where $v$ is its speed. The maximum height attained by the ball is

• A. $\frac{1}{2k}tan^{-1}\frac{ku^2}{g}$
• B. $\frac{1}{2k}ln\left(1+\frac{ku^2}{g}\right)$
• C. $\frac{1}{k}tan^{-1}\frac{ku^2}{2g}$
• D. $\frac{1}{k}ln\left(1+\frac{ku^2}{2g}\right)$

Answer 1

Answer: (B)

$\frac{1}{2k}\ln\left(1+\frac{ku^2}{g}\right)$

Answer 2

The net force acting on the ball when moving upward is

$$F = -mg - m k v^2.$$

Using $v \frac{dv}{dx} = \frac{dv}{dt}$, we have:

$$v \frac{dv}{dx} = -g - k v^2.$$

Rearrange to separate the variables:

$$\frac{v\,dv}{g + k v^2} = -dx.$$

For maximum height $H$, integrate from initial speed $u$ (at $x = 0$) to velocity $0$ (at $x = H$):

$$H = \int_0^H dx = \int_u^0 \frac{-v\,dv}{g+k v^2}.$$

Changing the limits we get:

$$H = \int_0^u \frac{v\,dv}{g+k v^2}.$$

Now, substitute $w = g + k v^2$ so that $dw = 2k\,v\,dv$ which implies $v\,dv = \frac{dw}{2k}$. The limits change as follows:

• When $v=0$, $w=g$.
• When $v=u$, $w=g+ku^2$.

Thus,

$$H = \frac{1}{2k} \int_{g}^{g+ku^2} \frac{dw}{w} = \frac{1}{2k} \left[\ln w\right]_{g}^{g+ku^2} = \frac{1}{2k}\ln\left(\frac{g+ku^2}{g}\right).$$

This simplifies to:

$$H = \frac{1}{2k}\ln\left(1+\frac{ku^2}{g}\right).$$

By applying the chain rule $v \frac{dv}{dx} = -g - k v^2$ and integrating with the substitution $w = g+kv^2$, the maximum height is found to be $\frac{1}{2k}\ln\left(1+\frac{ku^2}{g}\right)$.