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Question Number 465 by shubham last updated on 09/Jan/15

A particle moves with a central acceration varies as the cube of the distance. if it be projected from an apse at distance a from the origin with a velocity which is (√(2 )) times the velocity for a circle of radius a. show that the equation of path is its rcosθ/(√2) = a

$${A}\:{particle}\:{moves}\:{with}\:{a}\:{central}\:{acceration}\:{varies}\:{as}\:{the}\:{cube}\:{of}\:{the}\:{distance}.\:{if}\:{it}\:{be}\:{projected}\:{from}\:{an}\:{apse}\:{at}\:{distance}\:{a}\:{from}\:{the}\:{origin}\:{with}\:{a}\:{velocity}\:{which}\:{is}\:\sqrt{\mathrm{2}\:}\:{times}\:{the}\:{velocity}\:{for}\:{a}\:{circle}\:{of}\:{radius}\:{a}.\:{show}\:{that}\:{the}\:{equation}\:{of}\:{path}\:{is}\:{its}\:{r}\mathrm{cos}\theta/\sqrt{\mathrm{2}}\:=\:{a} \\ $$

Commented by prakash jain last updated on 09/Jan/15

Reformatted A particle moves with a central acceration varies as the cube of the distance. if it be projected from an apse at distance a from the origin with a velocity which is (√(2 )) times the velocity for a circle of radius a. show that the equation of path is its rcosθ/(√2) = a

$$\mathrm{Reformatted} \\ $$$${A}\:{particle}\:{moves}\:{with}\:{a}\:{central} \\ $$$${acceration}\:{varies}\:{as}\:{the}\:{cube}\: \\ $$$${of}\:{the}\:{distance}.\:{if}\:{it}\:{be}\:{projected} \\ $$$${from}\:{an}\:{apse}\:{at}\:{distance}\:{a} \\ $$$${from}\:{the}\:{origin}\:{with}\:{a}\:{velocity} \\ $$$${which}\:{is}\:\sqrt{\mathrm{2}\:}\:{times}\:{the}\:{velocity} \\ $$$${for}\:{a}\:{circle}\:{of}\:{radius}\:{a}.\:{show}\:{that} \\ $$$${the}\:{equation}\:{of}\:{path}\:{is} \\ $$$${its}\:{r}\mathrm{cos}\theta/\sqrt{\mathrm{2}}\:=\:{a} \\ $$

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