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Question Number 91946 by Rio Michael last updated on 03/May/20
a particle is projected from a point at a height 3h metres above a horizontal play ground. the direction of the projectile makes an angle α with the horizontal through the point of projection. show that if th greatest height reached above the point lc projection is h metres, then the horizontal distance travelled by the particle before striking the plane is 6h cotα metres. Find the vertical and horizontal component of the speed of the particle just before it hits the ground.
$$\mathrm{a}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height}\:\mathrm{3}{h}\:\mathrm{metres}\:\mathrm{above}\:\mathrm{a}\:\mathrm{horizontal} \\ $$$$\mathrm{play}\:\mathrm{ground}.\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{projectile}\:\mathrm{makes}\:\mathrm{an}\:\mathrm{angle}\:\alpha\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{projection}.\:\:\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{th}\:\mathrm{greatest} \\ $$$$\mathrm{height}\:\mathrm{reached}\:\mathrm{above}\:\mathrm{the}\:\mathrm{point}\:\mathrm{lc}\:\mathrm{projection}\:\mathrm{is}\:{h}\:\mathrm{metres},\:\mathrm{then}\:\mathrm{the}\:\mathrm{horizontal} \\ $$$$\mathrm{distance}\:\mathrm{travelled}\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{before}\:\mathrm{striking}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{6}{h}\:\mathrm{cot}\alpha\:\mathrm{metres}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{vertical}\:\mathrm{and}\:\mathrm{horizontal}\:\mathrm{component}\:\mathrm{of}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{just} \\ $$$$\mathrm{before}\:\mathrm{it}\:\mathrm{hits}\:\mathrm{the}\:\mathrm{ground}. \\ $$
Commented by mr W last updated on 04/May/20
apply methods as in Q91655
$${apply}\:{methods}\:{as}\:{in}\:{Q}\mathrm{91655} \\ $$
Commented by Rio Michael last updated on 04/May/20
oh sir am sorry i will fix that next time. and thanks
$$\mathrm{oh}\:\mathrm{sir}\:\mathrm{am}\:\mathrm{sorry}\:\mathrm{i}\:\mathrm{will}\:\mathrm{fix}\:\mathrm{that}\:\mathrm{next}\:\mathrm{time}.\:\mathrm{and}\:\mathrm{thanks} \\ $$
Commented by Rio Michael last updated on 04/May/20
sir in this question, should i assume the initial velocity is u, and apply that method? It seems it is at a hiegt reason why applying the parabolic property method doesn′t work.
$$\mathrm{sir}\:\mathrm{in}\:\mathrm{this}\:\mathrm{question},\:\mathrm{should}\:\mathrm{i}\:\mathrm{assume} \\ $$$$\mathrm{the}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{is}\:{u},\:\mathrm{and}\:\mathrm{apply}\:\mathrm{that}\:\mathrm{method}? \\ $$$$\mathrm{It}\:\mathrm{seems}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{hiegt}\:\mathrm{reason}\:\mathrm{why}\:\mathrm{applying}\:\mathrm{the} \\ $$$$\mathrm{parabolic}\:\mathrm{property}\:\mathrm{method}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{work}. \\ $$
Answered by mr W last updated on 04/May/20
Method I
$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$
Commented by Rio Michael last updated on 04/May/20
sir i have some questions about this method of yours. − do you assume that the origin is at the vertex that is at the point h on your diagram? or at O? − also when using your expression (y_1 /y_2 ) = ((x_1 /x_2 ))^2 how do you choose your points? that is your coordinates (x_1 y_1 ) and (x_2 ,y_2 )?
$$\mathrm{sir}\:\mathrm{i}\:\mathrm{have}\:\mathrm{some}\:\mathrm{questions}\:\mathrm{about}\:\mathrm{this}\:\mathrm{method}\:\mathrm{of}\:\mathrm{yours}. \\ $$$$−\:\mathrm{do}\:\mathrm{you}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{is}\:\mathrm{at}\:\mathrm{the}\:\mathrm{vertex}\: \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:{h}\:\mathrm{on}\:\mathrm{your}\:\mathrm{diagram}?\:\mathrm{or}\:\mathrm{at}\:\mathrm{O}? \\ $$$$−\:\mathrm{also}\:\mathrm{when}\:\mathrm{using}\:\mathrm{your}\:\mathrm{expression}\:\frac{{y}_{\mathrm{1}} }{{y}_{\mathrm{2}} }\:=\:\left(\frac{{x}_{\mathrm{1}} }{{x}_{\mathrm{2}} }\right)^{\mathrm{2}} \:\mathrm{how}\:\mathrm{do}\: \\ $$$$\mathrm{you}\:\mathrm{choose}\:\mathrm{your}\:\mathrm{points}?\:\mathrm{that}\:\mathrm{is}\:\mathrm{your}\:\mathrm{coordinates}\:\left({x}_{\mathrm{1}} {y}_{\mathrm{1}} \right)\:\mathrm{and} \\ $$$$\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right)? \\ $$
Commented by mr W last updated on 04/May/20
Commented by mr W last updated on 04/May/20
the path of the particle is ABCD. B is the highest point. the parabola is symmetric about B. we can image the particle starts from O. OA is the mirror image of CD. tan α=((2h)/((AC)/2)) ⇒((AC)/2)=2h cot α ((3h+h)/h)=(((OB′)/(A′B′)))^2 =4 OB′=2A′B′=AC=4h cot α=B′D A′D=A′B′+B′D=((AC)/2)+B′D A′D=2h cot α+4h cot α=6h cot α ⇒proved tan β=((2(3h+h))/(B′D))=((8h)/(4h cot α))=2 tan α u_(Dy) =(√(2g(h+3h)))=2(√(2gh)) u_(Dx) =(v_(Dx) /(tan β))=((2(√(2gh)))/(2 tan α))=cot α (√(2gh))
$${the}\:{path}\:{of}\:{the}\:{particle}\:{is}\:{ABCD}. \\ $$$${B}\:{is}\:{the}\:{highest}\:{point}.\:{the}\:{parabola} \\ $$$${is}\:{symmetric}\:{about}\:{B}.\:{we}\:{can}\:{image} \\ $$$${the}\:{particle}\:{starts}\:{from}\:{O}.\:{OA}\:{is} \\ $$$${the}\:{mirror}\:{image}\:{of}\:\:{CD}. \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}{h}}{\frac{{AC}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{AC}}{\mathrm{2}}=\mathrm{2}{h}\:\mathrm{cot}\:\alpha \\ $$$$\frac{\mathrm{3}{h}+{h}}{{h}}=\left(\frac{{OB}'}{{A}'{B}'}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${OB}'=\mathrm{2}{A}'{B}'={AC}=\mathrm{4}{h}\:\mathrm{cot}\:\alpha={B}'{D} \\ $$$${A}'{D}={A}'{B}'+{B}'{D}=\frac{{AC}}{\mathrm{2}}+{B}'{D} \\ $$$${A}'{D}=\mathrm{2}{h}\:\mathrm{cot}\:\alpha+\mathrm{4}{h}\:\mathrm{cot}\:\alpha=\mathrm{6}{h}\:\mathrm{cot}\:\alpha \\ $$$$\Rightarrow{proved} \\ $$$$ \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{2}\left(\mathrm{3}{h}+{h}\right)}{{B}'{D}}=\frac{\mathrm{8}{h}}{\mathrm{4}{h}\:\mathrm{cot}\:\alpha}=\mathrm{2}\:\mathrm{tan}\:\alpha \\ $$$${u}_{{Dy}} =\sqrt{\mathrm{2}{g}\left({h}+\mathrm{3}{h}\right)}=\mathrm{2}\sqrt{\mathrm{2}{gh}} \\ $$$${u}_{{Dx}} =\frac{{v}_{{Dx}} }{\mathrm{tan}\:\beta}=\frac{\mathrm{2}\sqrt{\mathrm{2}{gh}}}{\mathrm{2}\:\mathrm{tan}\:\alpha}=\mathrm{cot}\:\alpha\:\sqrt{\mathrm{2}{gh}} \\ $$
Commented by mr W last updated on 04/May/20
Commented by Rio Michael last updated on 04/May/20
thanks again sir
$$\mathrm{thanks}\:\mathrm{again}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 06/Nov/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 04/May/20
Method II
$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$
Commented by mr W last updated on 04/May/20
Commented by mr W last updated on 04/May/20
x=u cos α t y=u sin α t−(1/2)gt^2 max. hight at t=t_1 : u sin α−gt_1 =0 ⇒t_1 =((u sin α)/g) y_B =u sin α t_1 −(1/2)gt_1 ^2 =h (u sin α−(1/2)g×((u sin α)/g))((u sin α)/g)=h ⇒u sin α=(√(2gh)) ...(i) ball hits the ground at t=t_2 : y_D =u sin α t_2 −(1/2)gt_2 ^2 =−3h gt_2 ^2 −2(√(2gh))t_2 −6h=0 ⇒t_2 =3(√((2h)/g)) x_D =u cos α t_2 =3(√((2h)/g)) u cos α ⇒u cos α=(x_D /(3(√((2h)/g)))) ...(ii) (i)/(ii): tan α=((6h)/x_D ) ⇒x_D =6h cot α ⇒proved u_(Dx) =u cos α=((6h cot α)/(3(√((2h)/g))))=cot α (√(2gh)) u_(Dy) =u sin α−gt_2 =(√(2gh))−3(√(2gh))=−2(√(2gh))
$${x}={u}\:\mathrm{cos}\:\alpha\:{t} \\ $$$${y}={u}\:\mathrm{sin}\:\alpha\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${max}.\:{hight}\:{at}\:{t}={t}_{\mathrm{1}} : \\ $$$${u}\:\mathrm{sin}\:\alpha−{gt}_{\mathrm{1}} =\mathrm{0}\:\Rightarrow{t}_{\mathrm{1}} =\frac{{u}\:\mathrm{sin}\:\alpha}{{g}} \\ $$$${y}_{{B}} ={u}\:\mathrm{sin}\:\alpha\:{t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{gt}_{\mathrm{1}} ^{\mathrm{2}} ={h} \\ $$$$\left({u}\:\mathrm{sin}\:\alpha−\frac{\mathrm{1}}{\mathrm{2}}{g}×\frac{{u}\:\mathrm{sin}\:\alpha}{{g}}\right)\frac{{u}\:\mathrm{sin}\:\alpha}{{g}}={h} \\ $$$$\Rightarrow{u}\:\mathrm{sin}\:\alpha=\sqrt{\mathrm{2}{gh}}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${ball}\:{hits}\:{the}\:{ground}\:{at}\:{t}={t}_{\mathrm{2}} : \\ $$$${y}_{{D}} ={u}\:\mathrm{sin}\:\alpha\:{t}_{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{gt}_{\mathrm{2}} ^{\mathrm{2}} =−\mathrm{3}{h} \\ $$$${gt}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}{gh}}{t}_{\mathrm{2}} −\mathrm{6}{h}=\mathrm{0} \\ $$$$\Rightarrow{t}_{\mathrm{2}} =\mathrm{3}\sqrt{\frac{\mathrm{2}{h}}{{g}}}\: \\ $$$${x}_{{D}} ={u}\:\mathrm{cos}\:\alpha\:{t}_{\mathrm{2}} =\mathrm{3}\sqrt{\frac{\mathrm{2}{h}}{{g}}}\:{u}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow{u}\:\mathrm{cos}\:\alpha=\frac{{x}_{{D}} }{\mathrm{3}\sqrt{\frac{\mathrm{2}{h}}{{g}}}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{6}{h}}{{x}_{{D}} } \\ $$$$\Rightarrow{x}_{{D}} =\mathrm{6}{h}\:\mathrm{cot}\:\alpha\:\:\Rightarrow{proved} \\ $$$$ \\ $$$${u}_{{Dx}} ={u}\:\mathrm{cos}\:\alpha=\frac{\mathrm{6}{h}\:\mathrm{cot}\:\alpha}{\mathrm{3}\sqrt{\frac{\mathrm{2}{h}}{{g}}}}=\mathrm{cot}\:\alpha\:\sqrt{\mathrm{2}{gh}} \\ $$$${u}_{{Dy}} ={u}\:\mathrm{sin}\:\alpha−{gt}_{\mathrm{2}} =\sqrt{\mathrm{2}{gh}}−\mathrm{3}\sqrt{\mathrm{2}{gh}}=−\mathrm{2}\sqrt{\mathrm{2}{gh}} \\ $$
Commented by mr W last updated on 04/May/20
the same result as with method I.
$${the}\:{same}\:{result}\:{as}\:{with}\:{method}\:{I}. \\ $$
Commented by Rio Michael last updated on 04/May/20
thank you so much sir
$$\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$

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