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49P

Expert-verifiedFound in: Page 35

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: A hot-air balloon is ascending at the rate of **** ${\mathbf{12}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$ and role="math" localid="1662959960903" ${\mathbf{80}}{\mathbf{m}}$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?**

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

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

**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**

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),

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{0}+\mathrm{at}$

$41.38\mathrm{m}/\mathrm{s}=-12\mathrm{m}/\mathrm{s}+9.8\mathrm{m}/{\mathrm{s}}^{2}\times \mathrm{t}\phantom{\rule{0ex}{0ex}}53.38\mathrm{m}/\mathrm{s}=9.8\mathrm{m}/{\mathrm{s}}^{2}\times \mathrm{t}\phantom{\rule{0ex}{0ex}}\mathrm{t}=\frac{53.38\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^{2}}\phantom{\rule{0ex}{0ex}}=5.4469\mathrm{sec}\phantom{\rule{0ex}{0ex}}\approx 5.45\mathrm{sec}$

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

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