Question Number 46482 by peter frank last updated on 27/Oct/18
Commented by peter frank last updated on 27/Oct/18
$$\mathrm{help} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Oct/18
![horizontal velocity of projection=vcosα vertical velocity of projection=vsinα eqn for horigontal motion x=(vcosα)t eqn for vdrtical motion y=(vsinα)t−(1/2)gt^2 there is no accelartion along horizontal direction accelaration (g) is acting vertically down after time t_0 the vertical velocity=0 0=vsinα−gt_0 t_0 =((vsinα)/g) total time of flight=2t_0 =((2vsinα)/g) Range=(vcosα)(((2vsinα)/g))=v^2 ×((2sinαcosα)/g)=((v^2 sin2α)/g) greatest height when vertical component of velocity is zero. 0^2 =(vsinα)^2 −2gH [formula v^2 =u^2 −2as] H=((v^2 sin^2 α)/(2g)) y=(vsinα)t−(1/2)gt^2 y=(vsinα)((x/(vcosα)))−(1/2)g((x/(vcosα)))^2 y=xtanα−(1/2)g((x^2 /(v^2 cos^2 α))) y=Ax−Bx^2 Bx^2 −Ax=−y x^2 −(A/B)x=−(y/B) x^2 −2.x.(A/(2B))+(A^2 /(4B^2 ))=−(y/B)+(A^2 /(4B^2 )) (x−(A/(2B)))^2 =(((A^2 −4yB)/(4B^2 ))) (x−(A/(2B)))^2 =(−4B)×(1/(4B^2 ))(y−(A^2 /(4B))) (x−(A/(2B)))^2 =(−4)×((1/(4B)))(y−(A^2 /(4B))) X^2 =−4PY focus(0,−P) x−(A/(2B))=0 x=(A/(2B)) y−(A^2 /(4B))=−P y−(A^2 /(4B))=−(1/(4B)) y=((A^2 −1)/(4B)) focus((A/(2B)),((A^2 −1)/(4B))) directrix Y=P y−(A^2 /(4B))=(1/(4B)) y=((A^2 +1)/(4B))](https://www.tinkutara.com/question/Q46535.png)
$${horizontal}\:{velocity}\:{of}\:{projection}={vcos}\alpha \\ $$$${vertical}\:{velocity}\:{of}\:{projection}={vsin}\alpha \\ $$$${eqn}\:{for}\:{horigontal}\:{motion} \\ $$$${x}=\left({vcos}\alpha\right){t} \\ $$$${eqn}\:{for}\:{vdrtical}\:{motion} \\ $$$${y}=\left({vsin}\alpha\right){t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${there}\:{is}\:{no}\:{accelartion}\:{along}\:{horizontal}\:{direction} \\ $$$${accelaration}\:\left({g}\right)\:{is}\:{acting}\:{vertically}\:{down} \\ $$$${after}\:{time}\:{t}_{\mathrm{0}} \:\:{the}\:{vertical}\:{velocity}=\mathrm{0} \\ $$$$\mathrm{0}={vsin}\alpha−{gt}_{\mathrm{0}} \:\:\:\:{t}_{\mathrm{0}} =\frac{{vsin}\alpha}{{g}} \\ $$$${total}\:{time}\:{of}\:{flight}=\mathrm{2}{t}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{vsin}\alpha}{{g}} \\ $$$${Range}=\left({vcos}\alpha\right)\left(\frac{\mathrm{2}{vsin}\alpha}{{g}}\right)={v}^{\mathrm{2}} ×\frac{\mathrm{2}{sin}\alpha{cos}\alpha}{{g}}=\frac{{v}^{\mathrm{2}} {sin}\mathrm{2}\alpha}{{g}} \\ $$$${greatest}\:{height}\:{when}\:{vertical}\:{component}\:{of} \\ $$$${velocity}\:{is}\:{zero}. \\ $$$$\mathrm{0}^{\mathrm{2}} =\left({vsin}\alpha\right)^{\mathrm{2}} −\mathrm{2}{g}\boldsymbol{{H}}\:\:\left[{formula}\:{v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}{as}\right] \\ $$$$\boldsymbol{{H}}=\frac{{v}^{\mathrm{2}} {sin}^{\mathrm{2}} \alpha}{\mathrm{2}{g}} \\ $$$${y}=\left({vsin}\alpha\right){t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${y}=\left({vsin}\alpha\right)\left(\frac{{x}}{{vcos}\alpha}\right)−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{{x}}{{vcos}\alpha}\right)^{\mathrm{2}} \\ $$$${y}={xtan}\alpha−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{{x}^{\mathrm{2}} }{{v}^{\mathrm{2}} {cos}^{\mathrm{2}} \alpha}\right) \\ $$$$ \\ $$$$ \\ $$$${y}={Ax}−{Bx}^{\mathrm{2}} \\ $$$${Bx}^{\mathrm{2}} −{Ax}=−{y} \\ $$$${x}^{\mathrm{2}} −\frac{{A}}{{B}}{x}=−\frac{{y}}{{B}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}.{x}.\frac{{A}}{\mathrm{2}{B}}+\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}^{\mathrm{2}} }=−\frac{{y}}{{B}}+\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}^{\mathrm{2}} } \\ $$$$\left({x}−\frac{{A}}{\mathrm{2}{B}}\right)^{\mathrm{2}} =\left(\frac{{A}^{\mathrm{2}} −\mathrm{4}{yB}}{\mathrm{4}{B}^{\mathrm{2}} }\right) \\ $$$$\left({x}−\frac{{A}}{\mathrm{2}{B}}\right)^{\mathrm{2}} =\left(−\mathrm{4}{B}\right)×\frac{\mathrm{1}}{\mathrm{4}{B}^{\mathrm{2}} }\left({y}−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}}\right) \\ $$$$\left({x}−\frac{{A}}{\mathrm{2}{B}}\right)^{\mathrm{2}} =\left(−\mathrm{4}\right)×\left(\frac{\mathrm{1}}{\mathrm{4}{B}}\right)\left({y}−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}}\right) \\ $$$${X}^{\mathrm{2}} =−\mathrm{4}{PY} \\ $$$${focus}\left(\mathrm{0},−{P}\right) \\ $$$${x}−\frac{{A}}{\mathrm{2}{B}}=\mathrm{0}\:\:\:\:{x}=\frac{{A}}{\mathrm{2}{B}}\:\:\: \\ $$$${y}−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}}=−{P} \\ $$$${y}−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}}=−\frac{\mathrm{1}}{\mathrm{4}{B}} \\ $$$${y}=\frac{{A}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{B}} \\ $$$${focus}\left(\frac{{A}}{\mathrm{2}{B}},\frac{{A}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{B}}\right) \\ $$$${directrix}\:\:{Y}={P} \\ $$$${y}−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}}=\frac{\mathrm{1}}{\mathrm{4}{B}} \\ $$$${y}=\frac{{A}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{B}} \\ $$$$ \\ $$$$ \\ $$
Commented by peter frank last updated on 28/Oct/18
$$\mathrm{how}\:\mathrm{do}\:\mathrm{i}\:\mathrm{get}\:\mathrm{focus}\:\mathrm{and}\:\mathrm{directrix} \\ $$
Commented by peter frank last updated on 28/Oct/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$