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Question Number 19774 by Tinkutara last updated on 15/Aug/17

$$\mathrm{A}\mathrm{balloon}\mathrm{is}\mathrm{ascending}\mathrm{vertically}\mathrm{with}$$$$\mathrm{an}\mathrm{acceleration}\mathrm{of}0.2{\mathrm{ms}}^{-2}.\mathrm{Two}\mathrm{stones}$$$$\mathrm{are}\mathrm{dropped}\mathrm{from}\mathrm{it}\mathrm{at}\mathrm{an}\mathrm{interval}\mathrm{of}2\mathrm{s}.$$$$\mathrm{The}\mathrm{distance}\mathrm{between}\mathrm{them}\mathrm{when}\mathrm{the}$$$$\mathrm{second}\mathrm{stone}\mathrm{dropped}\mathrm{is}(\mathrm{take}g=9.8$$$${\mathrm{ms}}^{-2})$$

Answered by ajfour last updated on 15/Aug/17

$$\mathrm{For}\mathrm{first}\mathrm{stone}:$$$${\mathrm{y}}_1=\mathrm{h}+\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2$$$$\mathrm{where}\mathrm{h}\mathrm{is}\mathrm{height}\mathrm{of}\mathrm{balloon}\mathrm{at}\mathrm{t}=0$$$$\mathrm{u}\mathrm{is}\mathrm{initial}\mathrm{velocity}\mathrm{of}\mathrm{balloon}\mathrm{at}\mathrm{t}=0$$$$\mathrm{For}\mathrm{second}\mathrm{stone}:$$$${\mathrm{y}}_2=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$$$$\mathrm{so}{\mathrm{y}}_2-{\mathrm{y}}_1=\frac{1}{2}(\mathrm{g}+\mathrm{a}){\mathrm{t}}^2$$$${\mathrm{y}}_{\mathrm{rel}}=\left(\frac{1}{2}\times 10\times 4\right)\mathrm{m}=20\mathrm{m}.$$

Commented by Tinkutara last updated on 15/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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