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Question Number 96764 by Rio Michael last updated on 04/Jun/20
A particle P of mass m, is projected vertically upward with a speed u from a point A, on horizontal ground. When P is at x above its initial position, its speed is v. The only forces acting on P is its weight and resistance mgkv^2 . where k is a positive constant. (a) Show that the greatest height reached is (1/(2gk)) ln(1 +ku^2 ). (b) show that the speed with which P returns to A is (u/( (√(1+ ku^2 )))) .
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{P}\:\mathrm{of}\:\mathrm{mass}\:{m},\:\mathrm{is}\:\mathrm{projected}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{speed}\:{u}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:{A},\:\mathrm{on}\:\mathrm{horizontal}\:\mathrm{ground}.\:\mathrm{When}\:\mathrm{P}\:\mathrm{is}\:\mathrm{at}\:{x}\:\mathrm{above} \\ $$$$\mathrm{its}\:\mathrm{initial}\:\mathrm{position},\:\mathrm{its}\:\mathrm{speed}\:\mathrm{is}\:{v}.\:\mathrm{The}\:\mathrm{only}\:\mathrm{forces}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{P}\:\mathrm{is} \\ $$$$\mathrm{its}\:\mathrm{weight}\:\mathrm{and}\:\mathrm{resistance}\:{m}\mathrm{g}{kv}^{\mathrm{2}} .\:\mathrm{where}\:{k}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{constant}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{reached}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2g}{k}}\:\mathrm{ln}\left(\mathrm{1}\:+{ku}^{\mathrm{2}} \right). \\ $$$$\left(\mathrm{b}\right)\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{P}\:\mathrm{returns}\:\mathrm{to}\:\mathrm{A}\:\mathrm{is}\:\frac{{u}}{\:\sqrt{\mathrm{1}+\:{ku}^{\mathrm{2}} }}\:. \\ $$
Commented by mr W last updated on 04/Jun/20
Commented by mr W last updated on 04/Jun/20
no further comment...
$${no}\:{further}\:{comment}… \\ $$
Commented by Rio Michael last updated on 05/Jun/20
sorry sir, but on my phone it doesn′t look that way. its normal.
$$\mathrm{sorry}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{on}\:\mathrm{my}\:\mathrm{phone}\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{look}\:\mathrm{that}\:\mathrm{way}. \\ $$$$\mathrm{its}\:\mathrm{normal}. \\ $$
Commented by mr W last updated on 06/Jun/20
you have a large screen smart phone. then it′s not possible for you to control the line length while typing.
$${you}\:{have}\:{a}\:{large}\:{screen}\:{smart}\:{phone}. \\ $$$${then}\:{it}'{s}\:{not}\:{possible}\:{for}\:{you}\:{to} \\ $$$${control}\:{the}\:{line}\:{length}\:{while}\:{typing}. \\ $$
Commented by Rio Michael last updated on 08/Jun/20
i′ll try decreasing it. thanks
$$\mathrm{i}'\mathrm{ll}\:\mathrm{try}\:\mathrm{decreasing}\:\mathrm{it}.\:\mathrm{thanks} \\ $$
Answered by mr W last updated on 04/Jun/20
upwards: ma=−mg−mgkv^2 a=(dv/dt)=−g(1+kv^2 ) v(dv/dh)=−g(1+kv^2 ) ((2vkdv)/(1+kv^2 ))=−2gkdh ((d(1+kv^2 ))/(1+kv^2 ))=−2gkdh ∫_u ^0 ((d(1+kv^2 ))/(1+kv^2 ))=−2gk∫_0 ^h_(max) dh [ln (1+kv^2 )]_u ^0 =−2gk[h]_0 ^h_(max) ⇒h_(max) =(1/(2gk))ln (1+ku^2 ) downwards: ma=mg−mgkv^2 a=(dv/dt)=g(1−kv^2 ) −v(dv/dh)=g(1−kv^2 ) ((−2vkdv)/(1−kv^2 ))=2gkdh ((d(1−kv^2 ))/(1−kv^2 ))=2gkdh ∫_0 ^v_A ((d(1−kv^2 ))/(1−kv^2 ))=2gk∫_h_(max) ^0 dh [ln (1−kv^2 )]_0 ^v_A =2gk[h]_h_(max) ^0 ln (1−kv_A ^2 )=−2gkh_(max) ln (1−kv_A ^2 )=−ln (1+ku^2 ) 1−kv_A ^2 =(1/(1+ku^2 )) v_A ^2 =(u^2 /(1+ku^2 )) ⇒v_A =(u/( (√(1+ku^2 ))))
$${upwards}: \\ $$$${ma}=−{mg}−{mgkv}^{\mathrm{2}} \\ $$$${a}=\frac{{dv}}{{dt}}=−{g}\left(\mathrm{1}+{kv}^{\mathrm{2}} \right) \\ $$$${v}\frac{{dv}}{{dh}}=−{g}\left(\mathrm{1}+{kv}^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{2}{vkdv}}{\mathrm{1}+{kv}^{\mathrm{2}} }=−\mathrm{2}{gkdh} \\ $$$$\frac{{d}\left(\mathrm{1}+{kv}^{\mathrm{2}} \right)}{\mathrm{1}+{kv}^{\mathrm{2}} }=−\mathrm{2}{gkdh} \\ $$$$\int_{{u}} ^{\mathrm{0}} \frac{{d}\left(\mathrm{1}+{kv}^{\mathrm{2}} \right)}{\mathrm{1}+{kv}^{\mathrm{2}} }=−\mathrm{2}{gk}\int_{\mathrm{0}} ^{{h}_{{max}} } {dh} \\ $$$$\left[\mathrm{ln}\:\left(\mathrm{1}+{kv}^{\mathrm{2}} \right)\right]_{{u}} ^{\mathrm{0}} =−\mathrm{2}{gk}\left[{h}\right]_{\mathrm{0}} ^{{h}_{{max}} } \\ $$$$\Rightarrow{h}_{{max}} =\frac{\mathrm{1}}{\mathrm{2}{gk}}\mathrm{ln}\:\left(\mathrm{1}+{ku}^{\mathrm{2}} \right) \\ $$$$ \\ $$$${downwards}: \\ $$$${ma}={mg}−{mgkv}^{\mathrm{2}} \\ $$$${a}=\frac{{dv}}{{dt}}={g}\left(\mathrm{1}−{kv}^{\mathrm{2}} \right) \\ $$$$−{v}\frac{{dv}}{{dh}}={g}\left(\mathrm{1}−{kv}^{\mathrm{2}} \right) \\ $$$$\frac{−\mathrm{2}{vkdv}}{\mathrm{1}−{kv}^{\mathrm{2}} }=\mathrm{2}{gkdh} \\ $$$$\frac{{d}\left(\mathrm{1}−{kv}^{\mathrm{2}} \right)}{\mathrm{1}−{kv}^{\mathrm{2}} }=\mathrm{2}{gkdh} \\ $$$$\int_{\mathrm{0}} ^{{v}_{{A}} } \frac{{d}\left(\mathrm{1}−{kv}^{\mathrm{2}} \right)}{\mathrm{1}−{kv}^{\mathrm{2}} }=\mathrm{2}{gk}\int_{{h}_{{max}} } ^{\mathrm{0}} {dh} \\ $$$$\left[\mathrm{ln}\:\left(\mathrm{1}−{kv}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{{v}_{{A}} } =\mathrm{2}{gk}\left[{h}\right]_{{h}_{{max}} } ^{\mathrm{0}} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}−{kv}_{{A}} ^{\mathrm{2}} \right)=−\mathrm{2}{gkh}_{{max}} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}−{kv}_{{A}} ^{\mathrm{2}} \right)=−\mathrm{ln}\:\left(\mathrm{1}+{ku}^{\mathrm{2}} \right) \\ $$$$\mathrm{1}−{kv}_{{A}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}+{ku}^{\mathrm{2}} } \\ $$$${v}_{{A}} ^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} }{\mathrm{1}+{ku}^{\mathrm{2}} } \\ $$$$\Rightarrow{v}_{{A}} =\frac{{u}}{\:\sqrt{\mathrm{1}+{ku}^{\mathrm{2}} }} \\ $$
Commented by peter frank last updated on 04/Jun/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Rio Michael last updated on 05/Jun/20
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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