Motion in a Plane

JEE Physics · 62 questions · Page 7 of 7 · Click an option or "Show Solution" to reveal answer

Q61

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2 . If it were launched at an angle

\theta

0 with speed v0 then (g = 10 ms–2) :

{\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)

and

{v_0} = {5 \over 3}

ms-1

{\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)

and

{v_0} = {3 \over 5}

ms-1

{\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)

and

{v_0} = {3 \over 5}

ms-1

{\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)

and

{v_0} = {5 \over 3}

ms-1

Correct Answer

Option A

Solution

Equation of trajectory is given as y = 2x – 9x2 …(A) Comparing with equation:

y = x\tan \theta - {g \over {2{u^2}{{\cos }^2}\theta }}{x^2}

...(B) We get,

\tan \theta = 2

\therefore \cos \theta = {1 \over {\sqrt 5 }}

Also,

{g \over {2{u^2}{{\cos }^2}\theta }} = 9

\Rightarrow {{10} \over {2 \times 9 \times {{\left( {{1 \over {\sqrt 5 }}} \right)}^2}}} = {u^2};\,\,{u^2} = {{25} \over 9}

\Rightarrow u = {5 \over 3}m/s

Q62

A particle is moving with a velocity

\overrightarrow v \, = K(y\widehat i + x\widehat j),

where K is a constant. The general equation for its path is :

A y = x2 + constant

B y2 = x + constant

C y2 = x2 + constant

D xy = constant

Correct Answer

Option C

Solution

Given,

\overrightarrow v = K\left( {y\widehat i + x\widehat j} \right)

$\therefore$ Velocity in x direction,

{v_x} = {{dx} \over {dt}} = Ky\,

. . . . . (1) Velocity in y direction, vy =

\,{{dy} \over {dt}}

= Kx . . . . . . . (2) $\therefore$

{{\,{{dy} \over {dt}}} \over {{{dx} \over {dt}}}} = {{Kx} \over {Ky}}

$\Rightarrow$

{{dy} \over {dx}} = {x \over y}

$\Rightarrow$ ydy $=$ xdx Integrating both sides we get,

\int {ydx} = \int {xdx}

$\Rightarrow$

{{{y^2}} \over 2} = {{{x^2}} \over 2} + c

$\Rightarrow$

{y^2} = {x^2} + 2c

$\therefore$ General equation, y2 = x2 + constant.

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