Question Number 91196 by ajfour last updated on 28/Apr/20
Commented by ajfour last updated on 28/Apr/20
$${The}\:{ball}\:{has}\:{to}\:{be}\:{dropped}\:{on} \\ $$$${inciline}\:{so}\:{that}\:{it}\:{lands}\:{into} \\ $$$${the}\:{box}\:{and}\:{y}_{{P}} \:{be}\:{minimum}. \\ $$$${Find}\:{x}_{{P}} \:{in}\:{terms}\:{of}\:{b}. \\ $$
Answered by mr W last updated on 28/Apr/20
Commented by mr W last updated on 28/Apr/20
$${let}\:{x}_{{P}} =−{a} \\ $$$${u}=\sqrt{\mathrm{2}{g}\left({y}_{{P}} −{a}\right)} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{{a}+{b}}{{u}}\right)^{\mathrm{2}} =\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{4}\left({y}_{{P}} −{a}\right)} \\ $$$${y}_{{P}} ={a}+\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{4}{a}} \\ $$$$\frac{{dy}_{{P}} }{{da}}=\mathrm{1}+\frac{\left({a}+{b}\right)}{\mathrm{2}{a}}−\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{5}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{{b}}{\sqrt{\mathrm{5}}} \\ $$$${i}.{e}.\:{x}_{{P}} =−\frac{{b}}{\sqrt{\mathrm{5}}} \\ $$
Commented by ajfour last updated on 28/Apr/20
$${i}\:{think}\:{you}\:{got} \\ $$$${a}={b}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:\:,\:{Sir} \\ $$$$\left\{\left(\frac{{a}+{b}}{\mathrm{2}{a}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right\}^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 28/Apr/20
$${no}.\:{i}\:{think}\:{a}=\frac{{b}}{\sqrt{\mathrm{5}}}\:{is}\:{correct}\:{sir}. \\ $$$$\left\{\left(\frac{{a}+{b}}{\mathrm{2}{a}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\right\}^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$
Commented by ajfour last updated on 28/Apr/20
$${yes}\:{sir},\:{really},\:{and}\:{you}\: \\ $$$${have}\:{solved}\:{it}\:{very}\:{wisely},\:{Sir}! \\ $$