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STATEMENT-1-The-graph-between-kinetic-energy-and-vertical-displacement-is-a-straight-line-for-a-projectile-STATEMENT-2-The-graph-between-kinetic-energy-and-horizontal-displacement-is-a-straight-l




Question Number 19476 by Tinkutara last updated on 11/Aug/17
STATEMENT-1 : The graph between  kinetic energy and vertical displacement  is a straight line for a projectile.  STATEMENT-2 : The graph between  kinetic energy and horizontal  displacement is a straight line for a  projectile.  STATEMENT-3 : The graph between  kinetic energy and time is a parabola  for a projectile.
$$\mathrm{STATEMENT}-\mathrm{1}\::\:\mathrm{The}\:\mathrm{graph}\:\mathrm{between} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{and}\:\mathrm{vertical}\:\mathrm{displacement} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{for}\:\mathrm{a}\:\mathrm{projectile}. \\ $$$$\mathrm{STATEMENT}-\mathrm{2}\::\:\mathrm{The}\:\mathrm{graph}\:\mathrm{between} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{and}\:\mathrm{horizontal} \\ $$$$\mathrm{displacement}\:\mathrm{is}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{for}\:\mathrm{a} \\ $$$$\mathrm{projectile}. \\ $$$$\mathrm{STATEMENT}-\mathrm{3}\::\:\mathrm{The}\:\mathrm{graph}\:\mathrm{between} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{and}\:\mathrm{time}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parabola} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{projectile}. \\ $$
Answered by ajfour last updated on 12/Aug/17
K=(1/2)mv^2 = (1/2)m(v_y ^2 +v_x ^2 )      =(1/2)m[v_(y0) ^2 −2gy+v_(x0) ^2 ]  ⇒  K-y graph is straight line  further, since y=x((v_(y0) /v_(x0) ))−((gx^2 )/(2u^2 ))[1+(v_(y0) ^2 /v_(x0) ^2 )]  ⇒  K-t graph is a parabola          K=(1/2)m[(v_(y0) −gt)^2 +v_(x0) ^2 ]  so    K-t graph is a parabola .   statements  (1) and (3) are true.
$$\mathrm{K}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{m}\left(\mathrm{v}_{\mathrm{y}} ^{\mathrm{2}} +\mathrm{v}_{\mathrm{x}} ^{\mathrm{2}} \right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{m}\left[\mathrm{v}_{\mathrm{y0}} ^{\mathrm{2}} −\mathrm{2gy}+\mathrm{v}_{\mathrm{x0}} ^{\mathrm{2}} \right] \\ $$$$\Rightarrow\:\:\mathrm{K}-\mathrm{y}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{further},\:\mathrm{since}\:\mathrm{y}=\mathrm{x}\left(\frac{\mathrm{v}_{\mathrm{y0}} }{\mathrm{v}_{\mathrm{x0}} }\right)−\frac{\mathrm{gx}^{\mathrm{2}} }{\mathrm{2u}^{\mathrm{2}} }\left[\mathrm{1}+\frac{\mathrm{v}_{\mathrm{y0}} ^{\mathrm{2}} }{\mathrm{v}_{\mathrm{x0}} ^{\mathrm{2}} }\right] \\ $$$$\Rightarrow\:\:\mathrm{K}-\mathrm{t}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parabola} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{K}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{m}\left[\left(\mathrm{v}_{\mathrm{y0}} −\mathrm{gt}\right)^{\mathrm{2}} +\mathrm{v}_{\mathrm{x0}} ^{\mathrm{2}} \right] \\ $$$$\mathrm{so}\:\:\:\:\mathrm{K}-\mathrm{t}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parabola}\:. \\ $$$$\:\mathrm{statements}\:\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{3}\right)\:\mathrm{are}\:\mathrm{true}. \\ $$
Commented by Tinkutara last updated on 12/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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