Menu Close

Question-72672




Question Number 72672 by mr W last updated on 31/Oct/19
Commented by mr W last updated on 31/Oct/19
find u_(min)  for the stone to hit the  ground behind the sphere.  (see also Q72563, with thanks to   ajfour sir for the orginal question)
$${find}\:{u}_{{min}} \:{for}\:{the}\:{stone}\:{to}\:{hit}\:{the} \\ $$$${ground}\:{behind}\:{the}\:{sphere}. \\ $$$$\left({see}\:{also}\:{Q}\mathrm{72563},\:{with}\:{thanks}\:{to}\:\right. \\ $$$$\left.{ajfour}\:{sir}\:{for}\:{the}\:{orginal}\:{question}\right) \\ $$
Commented by malwaan last updated on 31/Oct/19
very easy !
$$\boldsymbol{{very}}\:\boldsymbol{{easy}}\:! \\ $$
Answered by ajfour last updated on 31/Oct/19
y=H−Ax^2          x^2 +(y−R)^2 =R^2   y=((u^2 sin^2 θ)/(2g))−Ax^2           ....(i)  ((2u^2 sin θcos θ)/g)=2d          ...(ii)  x^2 +(((u^2 sin^2 θ)/(2g))−Ax^2 −R)^2 =R^2   ⇒  x^2 −2A(((u^2 sin^2 θ)/(2g))−R)x^2 +A^2 x^4       +(((u^2 sin^2 θ)/(2g))−R)^2 −R^2 =0  coeff. of x^2 =0   { ((further if   ((u^2 sin^2 θ)/(2g))=2R)),((stone grazes the sphere top)) :}  ⇒  2A(((u^2 sin^2 θ)/(2g))−R)=1     ...(iii)  for  x=d , y=0   eq. (i) then gives  Ad^2 =((u^2 sin^2 θ)/(2g))  Now using (iii) in here  d^2 =((u^2 sin^2 θ)/(2g))×2(((u^2 sin^2 θ)/(2g))−R)  And from (ii)  u^2 =((gd)/(sin θcos θ)) ⇒  ((u^2 sin^2 θ)/(2g))=((dtan θ)/2)  ⇒  d^2 = dtan θ(((dtan θ)/2)−R)     t= tan θ     dt^2 −2Rt−2d=0  ⇒  t=tan θ=((2R+(√(4R^2 +8d^2 )))/(2d))    θ=tan^(−1) ((R/d)+(√(((R/d))^2 +2)) )    u^2 =((gdtan θ)/(1+tan^2 θ))     u^2 =((gd(((2R+(√(4R^2 +8d^2 )))/(2d))))/(1+(((2R+(√(4R^2 +8d^2 )))/(2d)))^2 ))  .....
$${y}={H}−{Ax}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\left({y}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${y}=\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}−{Ax}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$$\frac{\mathrm{2}{u}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta}{{g}}=\mathrm{2}{d}\:\:\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$${x}^{\mathrm{2}} +\left(\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}−{Ax}^{\mathrm{2}} −{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{A}\left(\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}−{R}\right){x}^{\mathrm{2}} +{A}^{\mathrm{2}} {x}^{\mathrm{4}} \\ $$$$\:\:\:\:+\left(\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} =\mathrm{0} \\ $$$${coeff}.\:{of}\:{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\begin{cases}{{further}\:{if}\:\:\:\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}=\mathrm{2}{R}}\\{{stone}\:{grazes}\:{the}\:{sphere}\:{top}}\end{cases} \\ $$$$\Rightarrow\:\:\mathrm{2}{A}\left(\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}−{R}\right)=\mathrm{1}\:\:\:\:\:…\left({iii}\right) \\ $$$${for}\:\:{x}={d}\:,\:{y}=\mathrm{0}\:\:\:{eq}.\:\left({i}\right)\:{then}\:{gives} \\ $$$${Ad}^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}} \\ $$$${Now}\:{using}\:\left({iii}\right)\:{in}\:{here} \\ $$$${d}^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}×\mathrm{2}\left(\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}−{R}\right) \\ $$$${And}\:{from}\:\left({ii}\right) \\ $$$${u}^{\mathrm{2}} =\frac{{gd}}{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}\:\Rightarrow\:\:\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}{g}}=\frac{{d}\mathrm{tan}\:\theta}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{d}^{\mathrm{2}} =\:{d}\mathrm{tan}\:\theta\left(\frac{{d}\mathrm{tan}\:\theta}{\mathrm{2}}−{R}\right) \\ $$$$\:\:\:{t}=\:\mathrm{tan}\:\theta \\ $$$$\:\:\:{dt}^{\mathrm{2}} −\mathrm{2}{Rt}−\mathrm{2}{d}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}=\mathrm{tan}\:\theta=\frac{\mathrm{2}{R}+\sqrt{\mathrm{4}{R}^{\mathrm{2}} +\mathrm{8}{d}^{\mathrm{2}} }}{\mathrm{2}{d}} \\ $$$$\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{R}}{{d}}+\sqrt{\left(\frac{{R}}{{d}}\right)^{\mathrm{2}} +\mathrm{2}}\:\right) \\ $$$$\:\:{u}^{\mathrm{2}} =\frac{{gd}\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta} \\ $$$$\:\:\:{u}^{\mathrm{2}} =\frac{{gd}\left(\frac{\mathrm{2}{R}+\sqrt{\mathrm{4}{R}^{\mathrm{2}} +\mathrm{8}{d}^{\mathrm{2}} }}{\mathrm{2}{d}}\right)}{\mathrm{1}+\left(\frac{\mathrm{2}{R}+\sqrt{\mathrm{4}{R}^{\mathrm{2}} +\mathrm{8}{d}^{\mathrm{2}} }}{\mathrm{2}{d}}\right)^{\mathrm{2}} } \\ $$$$….. \\ $$
Answered by mr W last updated on 31/Oct/19
Commented by mr W last updated on 31/Oct/19
there are many different cases. what  the image shows is the situation when  the sphere is far away. if the sphere  is near from the origin, it could look  differently. my method can not  treat such cases correctly. your solution  is for such cases then.
$${there}\:{are}\:{many}\:{different}\:{cases}.\:{what} \\ $$$${the}\:{image}\:{shows}\:{is}\:{the}\:{situation}\:{when} \\ $$$${the}\:{sphere}\:{is}\:{far}\:{away}.\:{if}\:{the}\:{sphere} \\ $$$${is}\:{near}\:{from}\:{the}\:{origin},\:{it}\:{could}\:{look} \\ $$$${differently}.\:{my}\:{method}\:{can}\:{not} \\ $$$${treat}\:{such}\:{cases}\:{correctly}.\:{your}\:{solution} \\ $$$${is}\:{for}\:{such}\:{cases}\:{then}. \\ $$
Commented by mr W last updated on 31/Oct/19
x=u cos θ t  y=u sin θ t−((gt^2 )/2)  y=x tan θ−(g/(2u^2 ))(1+tan^2  θ)x^2   let t=tan θ  y=tx−(g/(2u^2 ))(1+t^2 )x^2   y′=t−((g(1+t^2 )x)/u^2 )  B(d+R sin ϕ, R(1+cos ϕ))  −tan ϕ=t−((g(1+t^2 )(d+R sin ϕ))/u^2 )  ⇒((g(1+t^2 )(d+R sin ϕ))/u^2 )=t+tan ϕ  R(1+cos ϕ)=t(d+R sin ϕ)−(g/(2u^2 ))(1+t^2 )(d+R sin ϕ)^2   R(1+cos ϕ)=t(d+R sin ϕ)−(((d+R sin ϕ)(t+tan ϕ))/2)  ⇒t=tan ϕ+((2(1+cos ϕ))/(δ+sin ϕ)) with δ=(d/R)  ⇒((gR(1+t^2 )(δ+sin ϕ))/u^2 )=t+tan ϕ  ⇒((gR)/u^2 )=((t+tan ϕ)/((δ+sin ϕ)(1+t^2 )))  ⇒((gR)/(2u^2 ))=U=((1+cos ϕ+(δ+sin ϕ)tan ϕ)/((δ+sin ϕ)^2 +[(δ+sin ϕ)tan ϕ+2(1+cos ϕ)]^2 ))  (dU/dϕ)=0 .....
$${x}={u}\:\mathrm{cos}\:\theta\:{t} \\ $$$${y}={u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${y}={x}\:\mathrm{tan}\:\theta−\frac{{g}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right){x}^{\mathrm{2}} \\ $$$${let}\:{t}=\mathrm{tan}\:\theta \\ $$$${y}={tx}−\frac{{g}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right){x}^{\mathrm{2}} \\ $$$${y}'={t}−\frac{{g}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){x}}{{u}^{\mathrm{2}} } \\ $$$${B}\left({d}+{R}\:\mathrm{sin}\:\varphi,\:{R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)\right) \\ $$$$−\mathrm{tan}\:\varphi={t}−\frac{{g}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({d}+{R}\:\mathrm{sin}\:\varphi\right)}{{u}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{g}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({d}+{R}\:\mathrm{sin}\:\varphi\right)}{{u}^{\mathrm{2}} }={t}+\mathrm{tan}\:\varphi \\ $$$${R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)={t}\left({d}+{R}\:\mathrm{sin}\:\varphi\right)−\frac{{g}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left({d}+{R}\:\mathrm{sin}\:\varphi\right)^{\mathrm{2}} \\ $$$${R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)={t}\left({d}+{R}\:\mathrm{sin}\:\varphi\right)−\frac{\left({d}+{R}\:\mathrm{sin}\:\varphi\right)\left({t}+\mathrm{tan}\:\varphi\right)}{\mathrm{2}} \\ $$$$\Rightarrow{t}=\mathrm{tan}\:\varphi+\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)}{\delta+\mathrm{sin}\:\varphi}\:{with}\:\delta=\frac{{d}}{{R}} \\ $$$$\Rightarrow\frac{{gR}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\delta+\mathrm{sin}\:\varphi\right)}{{u}^{\mathrm{2}} }={t}+\mathrm{tan}\:\varphi \\ $$$$\Rightarrow\frac{{gR}}{{u}^{\mathrm{2}} }=\frac{{t}+\mathrm{tan}\:\varphi}{\left(\delta+\mathrm{sin}\:\varphi\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} }={U}=\frac{\mathrm{1}+\mathrm{cos}\:\varphi+\left(\delta+\mathrm{sin}\:\varphi\right)\mathrm{tan}\:\varphi}{\left(\delta+\mathrm{sin}\:\varphi\right)^{\mathrm{2}} +\left[\left(\delta+\mathrm{sin}\:\varphi\right)\mathrm{tan}\:\varphi+\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)\right]^{\mathrm{2}} } \\ $$$$\frac{{dU}}{{d}\varphi}=\mathrm{0}\:….. \\ $$
Commented by mr W last updated on 31/Oct/19
Commented by ajfour last updated on 31/Oct/19
should it be this way, i was   mistaken then sir, i thought  the trajectory would be envelop  the sphere symmetrically.
$${should}\:{it}\:{be}\:{this}\:{way},\:{i}\:{was}\: \\ $$$${mistaken}\:{then}\:{sir},\:{i}\:{thought} \\ $$$${the}\:{trajectory}\:{would}\:{be}\:{envelop} \\ $$$${the}\:{sphere}\:{symmetrically}. \\ $$
Commented by mr W last updated on 31/Oct/19

Leave a Reply

Your email address will not be published. Required fields are marked *