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-Dominant- [34]
3 years ago
10

lance wants to increase the range of motion in his hips. which of the best cool down strategy for reaching that goal using stati

c stretching? A. quickly swinging arms back and forth for more stress B. intensely holding stretched for five to 10 secs C. slowly increasing the amount of time spent stretching D.gradually extending stretches untill the muscle hurts
Health
2 answers:
Ray Of Light [21]3 years ago
6 0
C.) Slowly increasing the amount of time spent stretching. 

<span>For anything gradually increasing the difficulty or duration of the exercise always gives you the best chance of reaching the goal.</span>
stealth61 [152]3 years ago
5 0
My answer would be C.) Slowly increasing the amount of time spent stretching. 

Because if you do a good amount of stretching per day, while gradually increasing the time spent stretching, then you are bound to become more flexible around the hips :)
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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 \\  \\

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\bf \: ⇰ \: Speed \: of \: Jet  = 720 \: km/hr\\

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\bf \:➥ Time  \: taken \: to \: cover \: the \: distance  \: AB=15  \: sec  \\  \\

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

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

{ \bf \: ⇰  \: Let \: OL} = x   \:  \bf m \\  \\

\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]}

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

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\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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