A balloon starts rising from the surface of the Earth. The ascension rate is con

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A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to v_o. Due to the wind the balloon gathers the horizontal velocity component v_x=ay, where a is a constant and y is the height of ascent. Find the total, tangential, and normal accelerations of the balloon.

Options:

A . ar=av0;at=av0√1+(ayv0)2;aN=av0

B . ar=av0;at=av0;aN=0

C . ar=av0;at=a2y√1+a2y2v20;aN=av0 ⎷(1+a2y2v20)

D . ar=a2y√1+a2y2v20;at=av0;aN=av0

Answer: Option C
:
C
The path of the balloon will look something like this

A Balloon Starts Rising From The Surface Of The Earth. The A...

After t sec ballon would have gone a height of

v_{0} t

then at that very instance the balloon's

v_{x}

will be

v_{x} = a v_{0} t \Rightarrow a_{x} = a v_{0}

V_{y} = v_{0}

a_{y} = 0

a_{r} = t o t a l a c c e l e r a t i o n = a v_{0} . . . . . . . (1)

v_{t a n g e n t i c a l}

is actually the resultant velocity

\Rightarrow v_{t} = \sqrt{v^{2} 0 + (a v_{0} t)^{2}}

\Rightarrow v_{t} = v_{0} \sqrt{1 + a^{2} t^{2}}

a_{t} = \frac{d v_{t}}{d t} = \frac{v_{0} a^{2} t}{\sqrt{1 + a^{2} t^{2}}} \Rightarrow \frac{a^{2} y}{\sqrt{1 + {(\frac{a y}{v_{0}})}^{2}}}

Now we know

a_{t}^{2} + a_{N}^{2} = a_{r}^{2}

\Rightarrow \frac{v_{0}^{2} a^{4} t^{2}}{(1 + a^{2} t^{2})} + a_{N}^{2} = a^{2} v_{0}^{2}

a_{N} = \sqrt{v_{0}^{2} a^{4} t^{2} - \frac{v_{0}^{2} a^{4} t^{2}}{(1 + a^{2} t^{2})}}

a_{N} = a v_{0} \sqrt{\frac{1 + a^{2} t^{2} - a^{2} t^{2}}{(1 + a^{2} t^{2})}}

a_{N} = \frac{a v_{0}}{\sqrt{1 + a^{2} t^{2}}}

\Rightarrow \frac{a v_{0}}{\sqrt{1 + {(\frac{a y}{v_{0}})}^{2}}}

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