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 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!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com