Question

Find how the normal acceleration of the balloon varies as a function of the height of ascent.

• A. ${\omega }_n=\frac{a{v}_0}{\sqrt{1+{\left(\frac{a_y}{v_0}\right)}^2}}$
• B. ${\omega }_n=\frac{2a{v}_0}{\sqrt{1+{\left(\frac{a_y}{v_0}\right)}^2}}$
• C. ${\omega }_n=\frac{3a{v}_0}{2\sqrt{1+{\left(\frac{a_y}{v_0}\right)}^2}}$
• D. ${\omega }_n=\frac{a{v}_0}{2\sqrt{1+{\left(\frac{a_y}{v_0}\right)}^2}}$

Answer 1

The correct option is A ${\omega }_n=\frac{a{v}_0}{\sqrt{1+{\left(\frac{a_y}{v_0}\right)}^2}}$

Time required to reach height $y$ is $t=\frac{y}{v_o}$
Also, $\frac{dx}{dt}=ay=a{v}_{o}t$ integrating it we get, ${\int }_0^{x}dx=a{v}_o{\int }_0^{t}tdt⟹x=\frac{a{v}_o{t}^2}{2}=\frac{a{y}^2}{2v_o}$

Thus balloon is following a parabolic path
Vertical acceleration $a_y=0$
Horizontal acceleration $a_x=\frac{d{v}_x}{dt}=\frac{d}{dt}\left(a{v}_{0}t\right)=a{v}_o$
Net acceleration $a_{net}=a{v}_o$

Normal (radial) acceleration $a_n=a{v}_o\sin \theta =\frac{a{v}_o{v}_o}{\sqrt{{v_o}^2+{\left(a_y\right)}^2}}=\frac{a{v}_o}{\sqrt{1+{\left(\frac{a_y}{v_o}\right)}^2}}$