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A-stone-is-projected-from-a-point-on-the-ground-in-such-a-direction-so-as-to-hit-a-bird-on-the-top-of-a-telegraph-post-of-height-h-and-then-attain-a-height-2h-above-the-ground-If-at-an-instant-of-p

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Question Number 18131 by Tinkutara last updated on 15/Jul/17
A stone is projected from a point on the ground in such a direction so as to hit a bird on the top of a telegraph post of height h, and then attain a height 2h above the ground. If, at an instant of projection, the bird were to fly away horizontal with a uniform speed, find the ratio of the horizontal velocities of the bird and the stone, if the stone still hits the bird.
$$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{hit}\:\mathrm{a} \\ $$$$\mathrm{bird}\:\mathrm{on}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{telegraph}\:\mathrm{post}\:\mathrm{of} \\ $$$$\mathrm{height}\:{h},\:\mathrm{and}\:\mathrm{then}\:\mathrm{attain}\:\mathrm{a}\:\mathrm{height}\:\mathrm{2}{h} \\ $$$$\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{If},\:\mathrm{at}\:\mathrm{an}\:\mathrm{instant}\:\mathrm{of} \\ $$$$\mathrm{projection},\:\mathrm{the}\:\mathrm{bird}\:\mathrm{were}\:\mathrm{to}\:\mathrm{fly}\:\mathrm{away} \\ $$$$\mathrm{horizontal}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{speed},\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{velocities}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{bird}\:\mathrm{and}\:\mathrm{the}\:\mathrm{stone},\:\mathrm{if}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{still} \\ $$$$\mathrm{hits}\:\mathrm{the}\:\mathrm{bird}. \\ $$
Commented by Tinkutara last updated on 15/Jul/17
Commented by ajfour last updated on 15/Jul/17
y=xtan θ−((gx^2 )/(2u^2 cos^2 θ)) let m=tan θ , θ the∠ of projection. x=d, and and x=R−d when y=h d is the distance to telegraph post from point of projection. R is the range of stone.So h=mx−((gx^2 )/2)(1+m^2 ) g(1+m^2 )x^2 −2mx+2h=0 roots are x=d and x=R−d ⇒ sum of roots=R=((2m)/(g(1+m^2 ))) product = d(R−d)=((2h)/(g(1+m^2 ))) ⇒((d(R−d))/R)=(h/m) ....(i) 2h=((u^2 sin^2 θ)/(2g)) and R=((2u^2 sin θcos θ)/g) dividing equations we get ((8h)/R)=m ⇒ (h/m)=(R/8) substituting for (h/m) in (i) ((d(R−d))/R)=(R/8) ⇒ (d/R)(1−(d/R))=(1/8) ((d/R))^2 −((d/R))+(1/8)=0 ⇒ ((d/R)−(1/2))^2 =(1/8) ⇒ R−2d=(R/( (√2))) R(1−(1/( (√2))))=2d let v be the bird velocity (v/(ucos θ))=((R−2d)/(R−d))=((1−((2d)/R))/(1−d/R)) =((1−(1−(1/( (√2)))))/(1−((1/2)−(1/(2(√2))))))=((((1/( (√2)))))/((1/2)(1+(1/( (√2)))))) =(2/( (√2)+1)) .
$$\mathrm{y}=\mathrm{xtan}\:\theta−\frac{\mathrm{gx}^{\mathrm{2}} }{\mathrm{2u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta} \\ $$$$\mathrm{let}\:\mathrm{m}=\mathrm{tan}\:\theta\:\:,\:\theta\:\mathrm{the}\angle\:\mathrm{of}\:\mathrm{projection}. \\ $$$$\mathrm{x}=\mathrm{d},\:\mathrm{and}\:\mathrm{and}\:\mathrm{x}=\mathrm{R}−\mathrm{d}\:\:\mathrm{when}\:\mathrm{y}=\mathrm{h} \\ $$$$\:\mathrm{d}\:\mathrm{is}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{to}\:\mathrm{telegraph}\:\mathrm{post} \\ $$$$\mathrm{from}\:\mathrm{point}\:\mathrm{of}\:\mathrm{projection}.\:\mathrm{R}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{range}\:\mathrm{of}\:\mathrm{stone}.\mathrm{So} \\ $$$$\mathrm{h}=\mathrm{mx}−\frac{\mathrm{gx}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right) \\ $$$$\mathrm{g}\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right)\mathrm{x}^{\mathrm{2}} −\mathrm{2mx}+\mathrm{2h}=\mathrm{0} \\ $$$$\mathrm{roots}\:\mathrm{are}\:\mathrm{x}=\mathrm{d}\:\:\mathrm{and}\:\:\mathrm{x}=\mathrm{R}−\mathrm{d} \\ $$$$\Rightarrow\:\mathrm{sum}\:\mathrm{of}\:\mathrm{roots}=\mathrm{R}=\frac{\mathrm{2m}}{\mathrm{g}\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right)} \\ $$$$\mathrm{product}\:=\:\mathrm{d}\left(\mathrm{R}−\mathrm{d}\right)=\frac{\mathrm{2h}}{\mathrm{g}\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\frac{\mathrm{d}\left(\mathrm{R}−\mathrm{d}\right)}{\mathrm{R}}=\frac{\mathrm{h}}{\mathrm{m}}\:\:\:\:\:\:\:….\left(\mathrm{i}\right) \\ $$$$\mathrm{2h}=\frac{\mathrm{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2g}}\:\:\mathrm{and}\:\mathrm{R}=\frac{\mathrm{2u}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{g}} \\ $$$$\mathrm{dividing}\:\mathrm{equations}\:\mathrm{we}\:\mathrm{get} \\ $$$$\:\:\:\frac{\mathrm{8h}}{\mathrm{R}}=\mathrm{m}\:\:\Rightarrow\:\:\:\:\frac{\mathrm{h}}{\mathrm{m}}=\frac{\mathrm{R}}{\mathrm{8}} \\ $$$$\mathrm{substituting}\:\mathrm{for}\:\frac{\mathrm{h}}{\mathrm{m}}\:\:\mathrm{in}\:\left(\mathrm{i}\right) \\ $$$$\frac{\mathrm{d}\left(\mathrm{R}−\mathrm{d}\right)}{\mathrm{R}}=\frac{\mathrm{R}}{\mathrm{8}} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{d}}{\mathrm{R}}\left(\mathrm{1}−\frac{\mathrm{d}}{\mathrm{R}}\right)=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\left(\frac{\mathrm{d}}{\mathrm{R}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{d}}{\mathrm{R}}\right)+\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\left(\frac{\mathrm{d}}{\mathrm{R}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}\:\:\:\Rightarrow\:\:\mathrm{R}−\mathrm{2d}=\frac{\mathrm{R}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{R}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{2d} \\ $$$$\mathrm{let}\:\mathrm{v}\:\mathrm{be}\:\mathrm{the}\:\mathrm{bird}\:\mathrm{velocity} \\ $$$$\frac{\mathrm{v}}{\mathrm{ucos}\:\theta}=\frac{\mathrm{R}−\mathrm{2d}}{\mathrm{R}−\mathrm{d}}=\frac{\mathrm{1}−\frac{\mathrm{2d}}{\mathrm{R}}}{\mathrm{1}−\mathrm{d}/\mathrm{R}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)}=\frac{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\:\:. \\ $$
Commented by Tinkutara last updated on 15/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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