Question Number 27033 by Tinkutara last updated on 01/Jan/18
Answered by mrW1 last updated on 01/Jan/18
$$\left(\frac{{L}}{\mathrm{2}{h}}\right)^{\mathrm{2}} =\frac{{H}}{{H}−{h}}=\frac{{H}}{{H}−\frac{\mathrm{3}{H}}{\mathrm{4}}}=\mathrm{4} \\ $$$$\Rightarrow{L}=\mathrm{4}{h} \\ $$
Commented by mrW1 last updated on 02/Jan/18
$${actually}\:{I}\:{used}\:{no}\:{physics}\:{formula}. \\ $$$${we}\:{know}\:{the}\:{projectile}\:{is}\:{a}\:{parabola}. \\ $$$${in}\:{the}\:{coordinate}\:{system}\:{shown} \\ $$$${below}\:{we}\:{have} \\ $$$$\frac{{y}_{\mathrm{2}} }{{y}_{\mathrm{1}} }=\left(\frac{{x}_{\mathrm{2}} }{{x}_{\mathrm{1}} }\right)^{\mathrm{2}} \\ $$$${y}_{\mathrm{1}} ={H}−{h}=\frac{{H}}{\mathrm{4}} \\ $$$${y}_{\mathrm{2}} ={H} \\ $$$${x}_{\mathrm{1}} ={h} \\ $$$${x}_{\mathrm{2}} =\frac{{L}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}=\left(\frac{{L}}{\mathrm{2}{h}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{L}=\mathrm{4}{h} \\ $$
Commented by mrW1 last updated on 02/Jan/18
Commented by Tinkutara last updated on 02/Jan/18
Thank you very much Sir! I got the answer.