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Question Number 15115 by Tinkutara last updated on 07/Jun/17

Drops are falling regularly from a water  tap at a height of 9 m from the ground.  The 4^(th)  drop is about to fall from the  tap when the 1^(st)  hit the ground. Find  the distance between 2^(nd)  and 3^(rd)  drop.

$$\mathrm{Drops}\:\mathrm{are}\:\mathrm{falling}\:\mathrm{regularly}\:\mathrm{from}\:\mathrm{a}\:\mathrm{water} \\ $$$$\mathrm{tap}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\mathrm{9}\:\mathrm{m}\:\mathrm{from}\:\mathrm{the}\:\mathrm{ground}. \\ $$$$\mathrm{The}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{drop}\:\mathrm{is}\:\mathrm{about}\:\mathrm{to}\:\mathrm{fall}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{tap}\:\mathrm{when}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{hit}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{and}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{drop}. \\ $$

Answered by mrW1 last updated on 07/Jun/17

9=(1/2)×10×t^2   ⇒t=(√(1.8))  s  Δt=(t/3)=(√(0.2)) s  Δh_(2−3) =(1/2)×10×[(2Δt)^2 −Δt^2 ]  =(1/2)×10×3×0.2=3 m    or:  H_1 =(1/2)g(3t)^2   H_2 =(1/2)g(2t)^2   H_3 =(1/2)g(t)^2   H_4 =0  ΔH_(2−3) =(1/2)g(4t^2 −t^2 )=(3/2)gt^2 =(1/3)×(1/2)g(3t)^2   =(1/3)H_1 =(9/3)=3 m

$$\mathrm{9}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{t}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{t}=\sqrt{\mathrm{1}.\mathrm{8}}\:\:\mathrm{s} \\ $$$$\Delta\mathrm{t}=\frac{\mathrm{t}}{\mathrm{3}}=\sqrt{\mathrm{0}.\mathrm{2}}\:\mathrm{s} \\ $$$$\Delta\mathrm{h}_{\mathrm{2}−\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\left[\left(\mathrm{2}\Delta\mathrm{t}\right)^{\mathrm{2}} −\Delta\mathrm{t}^{\mathrm{2}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{3}×\mathrm{0}.\mathrm{2}=\mathrm{3}\:\mathrm{m} \\ $$$$ \\ $$$$\mathrm{or}: \\ $$$$\mathrm{H}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{3t}\right)^{\mathrm{2}} \\ $$$$\mathrm{H}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{2t}\right)^{\mathrm{2}} \\ $$$$\mathrm{H}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{t}\right)^{\mathrm{2}} \\ $$$$\mathrm{H}_{\mathrm{4}} =\mathrm{0} \\ $$$$\Delta\mathrm{H}_{\mathrm{2}−\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{4t}^{\mathrm{2}} −\mathrm{t}^{\mathrm{2}} \right)=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{3t}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{H}_{\mathrm{1}} =\frac{\mathrm{9}}{\mathrm{3}}=\mathrm{3}\:\mathrm{m} \\ $$

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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