Question

water drops are falling from a tap at 1 second interval . find the distance between 3rd and 5th drop when 6th drop is just coming out from the tap !

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Answer 1

Explanation:

so the water is dropped at the interval of 1 second

the distance between the tap 5th drop is

$s = ut + \frac{1}{2} a {t}^{2} \\ s = \frac{1}{2} 9.8 \times {1}^{2} \\ s5th = 4.9m \\ this \: is \: the \: distance \: of \: 5th \: drop \: from \: the \: tap \\ s = \frac{1}{2} a {t}^{2} \: \: \: because \: initial \: velocity \: is \: 0 \\ s = \frac{1}{2} 9.8 \times {3}^{2} \\ s3rd = 4.9 \times 9 \\ s3rd = 44.1m \\ s3rd - s5th \: gives \: the \: distance \: between \: drops \\ s3rd - s5th = 44.1 - 4.9 = 39.2m$

Answer 2

Answer:

Explanation:

so the water is dropped at the interval of 1 second

the distance between the tap 5th drop is

$s = ut + \frac{1}{2} a {t}^{2} \\ s = \frac{1}{2} 9.8 \times {1}^{2} \\ s5th = 4.9m \\ this \: is \: the \: distance \: of \: 5th \: drop \: from \: the \: tap \\ s = \frac{1}{2} a {t}^{2} \: \: \: because \: initial \: velocity \: is \: 0 \\ s = \frac{1}{2} 9.8 \times {3}^{2} \\ s3rd = 4.9 \times 9 \\ s3rd = 44.1m \\ s3rd - s5th \: gives \: the \: distance \: between \: drops \\ s3rd - s5th = 44.1 - 4.9 = 39.2m$

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