Question

Find how the horizontal drift of the balloon $x$ depends on height of ascenty?
(A) $\left( {\dfrac{a}{{{v_0}}}} \right){y^2}$
(B) $\left( {\dfrac{{2a}}{{{v_0}}}} \right){y^2}$
(C) $\left( {\dfrac{{3a}}{{2{v_0}}}} \right){y^2}$
(D) $\left( {\dfrac{a}{{2{v_0}}}} \right){y^2}$

Answer 1

Consider a balloon having horizontal component of velocity as ${V_x} = ay$ where a is a constant and y is the height of ascent. Let the ascension rate be equal to ${V_0}$ .Now try to determine how the horizontal drift depends on the height of the ascent.

Complete step by step answer:
${V_x} = ay$ ………………(1)
(where${V_x}$ is the Horizontal component of velocity a is constant and y is the height of ascent)
Rate of ascension is given by the equation ${V_0} = \dfrac{{dy}}{{dt}}$ rearranging the above equation we get $dy = {V_0}dt$
Integrating both sides
Put this in equation 1
$\int {dy} = \int {{V_0}dt} = y = {V_0}t$
Now we know the horizontal component of velocity is given by the equation ${V_x} = ay$, we know that ${V_x}$ is the derivative of displacement in x axis with respect to time
${V_x} = \dfrac{{dx}}{{dt}} = ay$
Again by rearranging the above equation it can be written as $dx = aydt$
From equation number 1 we already got value of $y = {V_0}t$ ,Substitute it and then integrate both sides we get,
$dx = aydt = a{V_0}tdt$
$\int {dx = \int {a{V_0}tdt = a{V_0}\int {tdt = a{V_0}\dfrac{{{t^2}}}{2}} } }$
Since ${V_0}t = y$ we can substitute it in the equation then will get,
$x = a{V_0}\dfrac{{{t^2}}}{2} \\\ \therefore x= \left( {\dfrac{a}{{2{V_0}}}} \right){y^2}$

Hence our answer is option D.

Note: First try to find out the equation for horizontal component of velocity and ascension rate then try to integrate and obtain value of y using the value of y try to find value of x or horizontal drift by integrating the obtained equation. Ascension rate is the rate at which an object rises in gas or liquid. It determines how fast the object moves upward.