Question

A balloon starts rising from the surface of the earth. The ascension rate is constant and equal to $V_0$. Due to the wind the balloon gathers a horizontal velocity component $V_x=Ky$ where K is a constant and y is the height of ascent.
The tangential and normal acceleration of the balloon at a height of ascent equal to h is

• A. $a_t=\frac{k^2{v}_{0}h}{\sqrt{v_0^2+k^2{h}^2}};a_n=\frac{k{v}^2}{v_0^2+k^2{h}^2}$
• B. $a_t=\frac{2k^2{v}_{0}h}{\sqrt{v_0^2+k^2{h}^2}};a_n=\frac{2k{v}^2}{v_0^2+k^2{h}^2}$
• C. $a_t=\frac{k^2{v}_{0}h}{\sqrt{v_0^2+k^2{h}^2}};a_n=\frac{2k{v}^2}{v_0^2+k^2{h}^2}$
• D. $a_t=\frac{2k^2{v}_0^2}{\sqrt{v_0^2+k^2{h}^2}};a_n=\frac{2k{v}_0}{v_0^2+k^2{h}^2}$

Answer 1

The correct option is A $a_t=\frac{k^2{v}_{0}h}{\sqrt{v_0^2+k^2{h}^2}};a_n=\frac{k{v}^2}{v_0^2+k^2{h}^2}$

${\overrightarrow{a}}_{net}=a_x\hat{i}+a_y\hat{j}$
$a_x=\frac{d{v}_x}{dt}=k\frac{dy}{dt}$ (from equation (ii))
$a_y=\frac{d{v}_y}{dt}=0$
$\Rightarrow {\overrightarrow{a}}_{net}=k{v}_0\hat{i}$ [using equation (i) and (ii)]

Let $\theta$ be the angle between velocity vector and x - axis. This means $tan\theta$ is the slop of the trajectory of the balloon.
Hence, $tan\theta =\frac{dy}{dx}$
Since, $x=\frac{k{y}^2}{2v_0}$
$\Rightarrow \frac{dy}{dx}=\frac{v_0}{ky}$
$a_t=a_{net}cos\theta$
$a_n=a_{net}sin\theta$
Put $a_{net}$, $cos\theta$ and $sin\theta$ to get the result.