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3 years ago

Why will one's feet drop if the head is lifted in a back float position in swimming?

1 answer:

6 0

That is because there must be perfect equalibrium between the body. It's like the fosbury flop, which was a why of high jumping that would allow high jumpers to "cheat" physics and leave equalibrium for short periods of time. If something goes up, something else must go down, this is a general rule becuase if it isn't true you would start floating out of the water

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Which of the following BEST describes a body in equilibrium?

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$\textsf {\large{\underline { \underline {SOLUTION:-}}}}$

⇰Let O be the point of observation on the ground OX.

⇰Let A and B be the two positions of the jet.

$\bf➠Then, \angle XOA=60⁰ \: and \: \angle \: XOB=30⁰ \\ \\$

$\bf ➠Draw \: AL \perp \: OX \: and \: BM \perp \: OX \\ \\$

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = (200 \times 15)m = 3000m \\ \\$

$\bf \therefore LM=AB=3000m \\ \\$

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$\bf➥From \: right \: \triangle OLA, \: we \: have: \\$

$\bf \frac{OL}{AL} =cot \: 60⁰= \frac{1}{ \sqrt{3} } \implies \frac{x}{h} = \frac{1}{ \sqrt{3} } \\ \\$

$\bf \implies \: x = \frac{h}{ \sqrt{3} } \: \: \: \: \: \: \: ....(1) \\ \\$

$\bf \:➥ From \: right \: \triangle \: OMB,we \: have: \\$

$\bf\frac{OM}{BM} =cot30⁰= \sqrt{3} \\ \\$

$\bf \implies \: \frac{x + 3000}{h} = \sqrt{3} \\ \\$

$\bf[ \boxed{➯OM=OL+LM=OL+AB=(x+3000)m \: and \: BM=h \: m]}$

$\bf \implies \: x + 3000 = \sqrt{3} h \\$

$\bf \implies \: x = ( \sqrt{3} h - 3000) \: \: \: \: \: \: ....(2) \\$

⇰ Equating the value of x from (1) and (2),we get;

$\bf \: \sqrt{3} h - 3000 = \frac{h}{ \sqrt{3} } \\ \\$

$\implies \bf \: 3h - 3000 \sqrt{3} = h \\$

$\bf \implies \: 2h = (3000 \times \sqrt{3} ) = (3000 \times 1.762) \\$

$\bf \implies \: h = (3000 \times 0.866) = 2598 \\ \\$

$\bf \:➥ Hence \: ,the \: required \: height \: is \: \boxed{ \boxed{ \bf2598 \: m.}} \\ \\$

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