A balloon starts rising from the earth’s surface. 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 constant and y is the height of ascent. Find the dependence of the following quantities on y.

Since the balloon is ascending at a constant rate.
So,  $v_0=\frac{dy}{dt}\Rightarrow dy=v_{0}dt\Rightarrow y=v_{0}t$
Also, we have
$v_x=\frac{dx}{dt}=ky$
$\Rightarrow dx=kydt=k{v}_{0}tdt \{\because y=v_{0}t\}$   
Integrating, we get 
$x=k{v}_0\left(\frac{t^2}{2}\right)\Rightarrow x=k\frac{v_0}{2}{\left(\frac{y}{v_0}\right)}^2\Rightarrow x=\frac{1}{2}\frac{k{y}^2}{v_0}$
A)  For finding the tangential and normal accelerations, we require an expression for the velocity as a function of height y. so we have 
$v_y=v_{0}and{v}_x=ky$
$\Rightarrow v=\sqrt{v_x^2+v_y^2}=\sqrt{v_0^2+k^2{y}^2}$
Therefore tangential acceleration, 
$a_T=\frac{dv}{dt}=\frac{k^{2}y}{\sqrt{v_0^2+k^2{y}^2}}\frac{dy}{dt}=\frac{k^{2}y{v}_0}{\sqrt{v_0^2+k^2{y}^2}}$
$\Rightarrow a=\frac{d{v}_x}{dt}=k\frac{dy}{dt}=k{v}_0 \{\because \frac{d{v}_y}{dt}=0\}$      
B) To find the normal acceleration, since we know that 
$a^2=a_N^2+a_T^2$
$\Rightarrow a_N=\sqrt{a^2-a_T^2}=\frac{k{v}_0}{\sqrt{1+{\left(\frac{ky}{v_0}\right)}^2}}$
C) $R_C=\frac{v^2}{a_c}=\frac{v_0}{k}{[1+{\left(\frac{ky}{v_0}\right)}^2]}^{3/2}$  
D) Curvature  $\alpha \frac{1}{R_c}$
$Asy\uparrow$ increases, $R_c\uparrow ,Curvature\downarrow$