49P - Fundamentals Of Physics

Question

Question: A hot-air balloon is ascending at the rate of  $\mathbf{12}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$ and $\mathbf{80}\mathbf{m}$is  above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground?  (b) With what speed does it hit the ground?

Step-by-Step Solution

Answer

1. Time (t) taken by the package to reach the ground is 5.45 sec.
2. Velocity (v) at which the package will hit the ground is 41.38 m/s.

Step 1: Given information

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

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

Step 2: Understanding the concept

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.

Step 3: First and second kinematic equation

The velocity in kinematic equation is expressed as,

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

(ii) 41 m/s

 

Step 4: Determination of speed of package hits the ground

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

Step 5: Determination of time taken by package to hits the ground

Using equation (ii),

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

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

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