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

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

Commented by mrW1 last updated on 16/Jun/17

(1)

Commented by chux last updated on 16/Jun/17

please explain this sir .

Commented by mrW1 last updated on 16/Jun/17

let us say particle 1 starts Δ t prior to

particle 2 .

particle 1 :

x_{1} = u_{1} (t + Δ t)

y_{1} = v_{1} (t + Δ t) - \frac{1}{2} g {(t + Δ t)}^{2}

particle 2 :

x_{2} = u_{2} t

y_{2} = v_{2} t - \frac{1}{2} {gt}^{2}

x^{'} = Δ x = x_{2} - x_{1} = (u_{2} - u_{1}) t - u_{1} Δ t

\Rightarrow x^{'} = Δ ut - u_{1} Δ t \dots (i)

y^{'} = Δ y = y_{2} - y_{1} = (v_{2} - v_{1}) t - v_{1} Δ t - \frac{1}{2} g Δ t (2 t + Δ t)

\Rightarrow y^{'} = Δ vt - v_{1} Δ t - g Δ t (t + \frac{Δ t}{2}) \dots (ii)

with Δ u = u_{2} - u_{1} and Δ v = v_{2} - v_{1}

from (i) :

t = \frac{x^{'} + u_{1} Δ t}{Δ u}

substituting in (ii) :

y^{'} = Δ v (\frac{x^{'} + u_{1} Δ t}{Δ u}) - v_{1} Δ t - g Δ t (\frac{x^{'} + u_{1} Δ t}{Δ u} + \frac{Δ t}{2}) \dots (ii)

y^{'} = \frac{Δ v}{Δ u} x^{'} + \frac{Δ v Δ {tu}_{1}}{Δ u} - v_{1} Δ t - \frac{g Δ t}{Δ u} x^{'} - \frac{g Δ t^{2} u_{1}}{Δ u} - \frac{g Δ t^{2}}{2}

y^{'} = (\frac{Δ v - g Δ t}{Δ u}) x^{'} + (\frac{Δ v Δ {tu}_{1} - Δ u Δ {tv}_{1} - g Δ t^{2} u_{1}}{Δ u} - \frac{g Δ t^{2}}{2})

\Rightarrow 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 .

Commented by Tinkutara last updated on 16/Jun/17

Thanks Sir!

Commented by chux last updated on 16/Jun/17

thanks a lot sir .

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