Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the nozzle when the first drop strikes the floor.

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7.35 m

2.45 m

4.18 m

2.94 m

answer is C.

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Detailed Solution

Using $H= \frac{1}{2} {gt}^{2} \Rightarrow {2H/g=t}^{2} \Rightarrow \frac{9.8 \times 2}{9.8} = t^{2} \Rightarrow t = \sqrt{2} s$

For first drop, t = 0; For second drop, $t=Δt$

For third drop, $t=2Δt$

Let h be the distance travelled by second drop when third drop begins to fall $∴ h= \frac{1}{2} g {(\sqrt{2} - Δ t)}^{2} \Rightarrow 0 = \frac{1}{2} g {(\sqrt{2} - Δ t)}^{2} \Rightarrow Δ t = \frac{1}{\sqrt{2}}$

$∴ h= \frac{1}{2} g {(\sqrt{2} - \frac{1}{\sqrt{2}})}^{2} = \frac{1}{2} \times 9.8 \times \frac{1}{2} = \frac{9.8}{4} = 2.45 m$

Height of second drop from nozzle = 2.45 m

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