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 horizontal drift of the balloon depends on the height of the ascent as

• A. $\frac{K{y}^2}{V_0}$
• B. $\frac{K{y}^2}{2V_0}$
• C. $\frac{2K{y}^2}{V_0}$
• D. $\frac{K{y}^2}{\sqrt{2}{V}_0}$

Answer 1

The correct option is B $\frac{K{y}^2}{2V_0}$

$v_y=\frac{dy}{dt}=v_0$ ...(i)
$v_x=ky\Rightarrow \frac{dx}{dt}=ky$ ...(ii)
Dividing (i) and (ii), we get
$\frac{dy}{dx}=\frac{v_0}{ky}$
$\Rightarrow {\int }_o^{y}kydy={\int }_0^x{v}_{0}dx$
or, $k{y}^2=2v_{0}x$
or, $x=\frac{k{y}^2}{2v_0}$