A balloon is ascending vertically with an acceleration of 1 $\mathrm{m} \mathrm{s}^{-2}$. Two stones are dropped from it at an interval of $2 \mathrm{~s}$. Find the distance between them $1.5 \mathrm{~s}$ after the second stone is released.

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Aicharya · Accepted Answer

Let at any time $t=0$, the balloon be at position $A$, where its velocity is $u$. At $t=2 \mathrm{~s}$, it reaches $B, \mathrm{\pi}$ where its velocity becomes $v$, then $A B=S=u \times 2+\frac{1}{2} a(2)^{2}=2 u+2$ Also $v=u+a \times 2=u+2$, First stone: $-S_{1}=u \times 3.5+\frac{1}{2}(-g)(3.5)^{2}$ Second stone: $-S_{2}=v \times 1.5+\frac{1}{2}(-g)(1.5)^{2}$ Required distance between the stones $x=S_{1}+S-S_{2}$ Solve to get $x=55 \mathrm{~m}$. Alternatively: (This method can be understood in proper way after studying relative velocity.) If we work from the frame of balloon, then the acceleration of each stone w.r.t. balloon will be $g+a$ after releasing from it. The initial velocity of each stone will be zero w.r.t. balloon. $S_{1}=\frac{1}{2}(g+a)(3.5)^{2}, S_{2}=\frac{1}{2}(g+a)(1.5)^{2} ; x=S_{1}-S_{2}=55 \mathrm{~m}$ ,

A balloon is ascending vertically with an acceleration of 1 \(\mathrm{m} \mathrm{s}^{-2}\). Two stones are…

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