Answer
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Hint: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.
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.
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