Question

A balloon, going upwards with a velocity 12 m/s, is at a height of 65 m from the earth at any instant.
Exactly at this instant a packet drops from it. How much time will the packet take to reach the earth?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{height},\mathrm{h}=65\mathrm{m} \\ \mathrm{velocity},\mathrm{u}=-12\mathrm{m}/\mathrm{s}\mathrm{as}\mathrm{it}\mathrm{is}\mathrm{going}\mathrm{upward} \\ \mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity},\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{h}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2 \\ 65=-12\mathrm{t}+\frac{1}{2}\times 10{\mathrm{t}}^2 \\ 5{\mathrm{t}}^2-12\mathrm{t}-65=0 \\ 5{\mathrm{t}}^2-25\mathrm{t}+13\mathrm{t}-65=0 \\ 5\mathrm{t}\left(\mathrm{t}-5\right)+13\left(\mathrm{t}-5\right)=0 \\ \left(5\mathrm{t}+13\right)\left(\mathrm{t}-5\right)=0 \\ \mathrm{t}=-\frac{13}{5}\mathrm{s}\mathrm{or}\mathrm{t}=5\mathrm{s} \\ \mathrm{As}\mathrm{we}\mathrm{do}\mathrm{not}\mathrm{take}\mathrm{negative}\mathrm{time}.\mathrm{So}\mathrm{the}\mathrm{packet}\mathrm{will}\mathrm{take}5\mathrm{s}\mathrm{to}\mathrm{reach}\mathrm{the}\mathrm{earth}. \end{gathered}$$