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Path-of-a-projectile-as-seen-from-another-projectile-is-a-1-Straight-line-2-Parabola-3-Ellipse-4-Hyperbola-

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Question Number 15961 by Tinkutara last updated on 16/Jun/17
Path of a projectile as seen from another projectile is a (1) Straight line (2) Parabola (3) Ellipse (4) Hyperbola
$$\mathrm{Path}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile}\:\mathrm{as}\:\mathrm{seen}\:\mathrm{from}\:\mathrm{another} \\ $$$$\mathrm{projectile}\:\mathrm{is}\:\mathrm{a} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Straight}\:\mathrm{line} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Parabola} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Ellipse} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Hyperbola} \\ $$
Commented by mrW1 last updated on 16/Jun/17
(1)
$$\left(\mathrm{1}\right) \\ $$
Commented by chux last updated on 16/Jun/17
please explain this sir.
$$\mathrm{please}\:\mathrm{explain}\:\mathrm{this}\:\mathrm{sir}. \\ $$
Commented by mrW1 last updated on 16/Jun/17
let us say particle 1 starts Δt prior to particle 2. particle 1: x_1 =u_1 (t+Δt) y_1 =v_1 (t+Δt)−(1/2)g(t+Δt)^2 particle 2: x_2 =u_2 t y_2 =v_2 t−(1/2)gt^2 x′=Δx=x_2 −x_1 =(u_2 −u_1 )t−u_1 Δt ⇒x′=Δut−u_1 Δt ...(i) y′=Δy=y_2 −y_1 =(v_2 −v_1 )t−v_1 Δt−(1/2)gΔt(2t+Δt) ⇒y′=Δvt−v_1 Δt−gΔt(t+((Δt)/2)) ...(ii) with Δu=u_2 −u_1 and Δv=v_2 −v_1 from (i): t=((x′+u_1 Δt)/(Δu)) substituting in (ii): y′=Δv(((x′+u_1 Δt)/(Δu)))−v_1 Δt−gΔt(((x′+u_1 Δt)/(Δu))+((Δt)/2)) ...(ii) y′=((Δv)/(Δu))x′+((ΔvΔtu_1 )/(Δu))−v_1 Δt−((gΔt)/(Δu))x′−((gΔt^2 u_1 )/(Δu))−((gΔt^2 )/2) y′=(((Δv−gΔt)/(Δu)))x′+(((ΔvΔtu_1 −ΔuΔtv_1 −gΔt^2 u_1 )/(Δu))−((gΔt^2 )/2)) ⇒y′=cx′+d with c,d=constant that means upon t≥0, the path of particle 2 seen from particle 1 is a straight, no mater how different their velocities are and whether they start at same time or at different time.
$$\mathrm{let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{particle}\:\mathrm{1}\:\mathrm{starts}\:\Delta\mathrm{t}\:\mathrm{prior}\:\mathrm{to} \\ $$$$\mathrm{particle}\:\mathrm{2}.\: \\ $$$$ \\ $$$$\mathrm{particle}\:\mathrm{1}: \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{u}_{\mathrm{1}} \left(\mathrm{t}+\Delta\mathrm{t}\right) \\ $$$$\mathrm{y}_{\mathrm{1}} =\mathrm{v}_{\mathrm{1}} \left(\mathrm{t}+\Delta\mathrm{t}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{t}+\Delta\mathrm{t}\right)^{\mathrm{2}} \\ $$$$\mathrm{particle}\:\mathrm{2}: \\ $$$$\mathrm{x}_{\mathrm{2}} =\mathrm{u}_{\mathrm{2}} \mathrm{t} \\ $$$$\mathrm{y}_{\mathrm{2}} =\mathrm{v}_{\mathrm{2}} \mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{x}'=\Delta\mathrm{x}=\mathrm{x}_{\mathrm{2}} −\mathrm{x}_{\mathrm{1}} =\left(\mathrm{u}_{\mathrm{2}} −\mathrm{u}_{\mathrm{1}} \right)\mathrm{t}−\mathrm{u}_{\mathrm{1}} \Delta\mathrm{t} \\ $$$$\Rightarrow\mathrm{x}'=\Delta\mathrm{ut}−\mathrm{u}_{\mathrm{1}} \Delta\mathrm{t}\:\:\:\:\:…\left(\mathrm{i}\right) \\ $$$$\mathrm{y}'=\Delta\mathrm{y}=\mathrm{y}_{\mathrm{2}} −\mathrm{y}_{\mathrm{1}} =\left(\mathrm{v}_{\mathrm{2}} −\mathrm{v}_{\mathrm{1}} \right)\mathrm{t}−\mathrm{v}_{\mathrm{1}} \Delta\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\Delta\mathrm{t}\left(\mathrm{2t}+\Delta\mathrm{t}\right) \\ $$$$\Rightarrow\mathrm{y}'=\Delta\mathrm{vt}−\mathrm{v}_{\mathrm{1}} \Delta\mathrm{t}−\mathrm{g}\Delta\mathrm{t}\left(\mathrm{t}+\frac{\Delta\mathrm{t}}{\mathrm{2}}\right)\:\:…\left(\mathrm{ii}\right) \\ $$$$\mathrm{with}\:\Delta\mathrm{u}=\mathrm{u}_{\mathrm{2}} −\mathrm{u}_{\mathrm{1}} \:\mathrm{and}\:\Delta\mathrm{v}=\mathrm{v}_{\mathrm{2}} −\mathrm{v}_{\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{from}\:\left(\mathrm{i}\right): \\ $$$$\mathrm{t}=\frac{\mathrm{x}'+\mathrm{u}_{\mathrm{1}} \Delta\mathrm{t}}{\Delta\mathrm{u}} \\ $$$$\mathrm{substituting}\:\mathrm{in}\:\left(\mathrm{ii}\right): \\ $$$$\mathrm{y}'=\Delta\mathrm{v}\left(\frac{\mathrm{x}'+\mathrm{u}_{\mathrm{1}} \Delta\mathrm{t}}{\Delta\mathrm{u}}\right)−\mathrm{v}_{\mathrm{1}} \Delta\mathrm{t}−\mathrm{g}\Delta\mathrm{t}\left(\frac{\mathrm{x}'+\mathrm{u}_{\mathrm{1}} \Delta\mathrm{t}}{\Delta\mathrm{u}}+\frac{\Delta\mathrm{t}}{\mathrm{2}}\right)\:\:…\left(\mathrm{ii}\right) \\ $$$$\mathrm{y}'=\frac{\Delta\mathrm{v}}{\Delta\mathrm{u}}\mathrm{x}'+\frac{\Delta\mathrm{v}\Delta\mathrm{tu}_{\mathrm{1}} }{\Delta\mathrm{u}}−\mathrm{v}_{\mathrm{1}} \Delta\mathrm{t}−\frac{\mathrm{g}\Delta\mathrm{t}}{\Delta\mathrm{u}}\mathrm{x}'−\frac{\mathrm{g}\Delta\mathrm{t}^{\mathrm{2}} \mathrm{u}_{\mathrm{1}} }{\Delta\mathrm{u}}−\frac{\mathrm{g}\Delta\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\: \\ $$$$\mathrm{y}'=\left(\frac{\Delta\mathrm{v}−\mathrm{g}\Delta\mathrm{t}}{\Delta\mathrm{u}}\right)\mathrm{x}'+\left(\frac{\Delta\mathrm{v}\Delta\mathrm{tu}_{\mathrm{1}} −\Delta\mathrm{u}\Delta\mathrm{tv}_{\mathrm{1}} −\mathrm{g}\Delta\mathrm{t}^{\mathrm{2}} \mathrm{u}_{\mathrm{1}} }{\Delta\mathrm{u}}−\frac{\mathrm{g}\Delta\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right)\: \\ $$$$\Rightarrow\mathrm{y}'=\mathrm{cx}'+\mathrm{d}\:\mathrm{with}\:\mathrm{c},\mathrm{d}=\mathrm{constant} \\ $$$$ \\ $$$$\mathrm{that}\:\mathrm{means}\:\mathrm{upon}\:\mathrm{t}\geqslant\mathrm{0},\:\mathrm{the}\:\mathrm{path}\:\mathrm{of}\: \\ $$$$\mathrm{particle}\:\mathrm{2}\:\mathrm{seen}\:\mathrm{from}\:\mathrm{particle}\:\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{straight}, \\ $$$$\mathrm{no}\:\mathrm{mater}\:\mathrm{how}\:\mathrm{different}\:\mathrm{their}\:\mathrm{velocities} \\ $$$$\mathrm{are}\:\mathrm{and}\:\mathrm{whether}\:\mathrm{they}\:\mathrm{start}\:\mathrm{at}\:\mathrm{same}\:\mathrm{time} \\ $$$$\mathrm{or}\:\mathrm{at}\:\mathrm{different}\:\mathrm{time}. \\ $$
Commented by Tinkutara last updated on 16/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by chux last updated on 16/Jun/17
thanks a lot sir.
$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{sir}. \\ $$

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