Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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.

$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com