91Ó°ÊÓ

Initial speed for package, $\left({\mathrm{V}}_{∘}\right)=-12\mathrm{m}/\mathrm{s}$

Height from which package is ${{\mathrm{v}}_{\mathrm{f}}}^{\text{2}}=v_{\text{0}}^{\text{2}}\text{+ 2as}$dropped$\mathrm{s}=80\mathrm{m}$

Acceleration=Gravitational Acceleration$\mathrm{a}=9.8\mathrm{m}/{\mathrm{s}}^2$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the Kinematic equation, the velocity and time to reach the ground dropped from hot air balloon at height$\mathbf{80}\mathbf{}\mathbf{m}$ can be calculated.

The velocity in kinematic equation is expressed as,

(i) ${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\text{0}}\text{+ at}$

(ii) 41 m/s

Consider downward as positive and upward as negative.

Calculate velocity (Part b), which can then be used in the calculation of time.

The balloon is ascending at the rate of -12m/s, and the package is dropped over the side.

Using equation (i), the final velocity will be

${{\mathrm{v}}_{\mathrm{f}}}^2=1712{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}$

${\mathrm{v}}_{\mathrm{f}}=41.38\text{m/s}$

The velocity of the package when it hits the ground is 41.38 m/s

Using equation (ii),

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at}$

$$\begin{gathered} 41.38\mathrm{m}/\mathrm{s}=-12\mathrm{m}/\mathrm{s}+9.8\mathrm{m}/{\mathrm{s}}^2×\mathrm{t} \\ 53.38\mathrm{m}/\mathrm{s}=9.8\mathrm{m}/{\mathrm{s}}^2×\mathrm{t} \\ \mathrm{t}=\frac{53.38\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =5.4469\sec \\ ≈5.45\sec \end{gathered}$$

The time taken by the packet to reach the ground is 5.45sec.