Question

Find how the tangential acceleration ${\omega }_{\tau }$of the balloon depends on the height of ascent.

• A. ${\omega }_{\tau }=\frac{2a^{2}y}{\sqrt{1+{\left(\frac{ay}{v_0}\right)}^2}}$
• B. ${\omega }_{\tau }=\frac{a^{2}y}{\sqrt{1+{\left(\frac{ay}{v_0}\right)}^2}}$
• C. ${\omega }_{\tau }=\frac{a^{2}y}{\sqrt{1+{\left(\frac{2ay}{v_0}\right)}^2}}$
• D. ${\omega }_{\tau }=\frac{a^{2}y}{2\sqrt{1+{\left(\frac{ay}{v_0}\right)}^2}}$

Answer 1

Answer: (B)

${\omega }_{\tau }=\frac{a^{2}y}{\sqrt{1+{\left(\frac{ay}{v_0}\right)}^2}}$

We have $tan\theta =\frac{v_y}{v_x}=\frac{v_o}{ay}$
We have acceleration in the vertical direction as $a_y=0$
and
$a_x=\frac{d{v}_x}{dt}=a\frac{dy}{dt}=a{v}_y=a{v}_o$
Thus $a_{net}=a{v}_o$
So we get the tangential component of acceleration as
$a_t=a_{net}cos\theta =a{v}_o\times \frac{ay}{\sqrt{v_o^2+(ay{)}^2}}=\frac{a^{2}y}{\sqrt{1+(\frac{ay}{v_o}{)}^2}}$