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Motion in A Plane question

2019 · 12 Apr · Shift 1 · Q57
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Motion in A Plane question

2019 · 12 Apr · Shift 1 · Q57

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2 . If it were launched at an angle θ\thetaθ 0 with speed v0 then (g = 10 ms–2) :
  1. A
    θ0=cos⁡−1(15){\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)θ0​=cos−1(5​1​) and v0=53{v_0} = {5 \over 3}v0​=35​ ms-1
  2. B
    θ0=cos⁡−1(25){\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)θ0​=cos−1(5​2​) and v0=35{v_0} = {3 \over 5}v0​=53​ ms-1
  3. C
    θ0=sin⁡−1(25){\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)θ0​=sin−1(5​2​) and v0=35{v_0} = {3 \over 5}v0​=53​ ms-1
  4. D
    θ0=sin⁡−1(15){\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)θ0​=sin−1(5​1​) and v0=53{v_0} = {5 \over 3}v0​=35​ ms-1
View written solutionFree

Correct answer: A

  1. Use the standard equation of a projectile

For a projectile launched from the origin with speed v0v_0v0​ at angle θ0\theta_0θ0​, its trajectory is

y=xtan⁡θ0−gx22v02cos⁡2θ0y = x\tan\theta_0 - \frac{g x^2}{2v_0^2\cos^2\theta_0}y=xtanθ0​−2v02​cos2θ0​gx2​

Given in the question:

y=2x−9x2y = 2x - 9x^2y=2x−9x2

Also, g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2.

  1. Compare coefficients of xxx and x2x^2x2

From

y=xtan⁡θ0−gx22v02cos⁡2θ0y = x\tan\theta_0 - \frac{g x^2}{2v_0^2\cos^2\theta_0}y=xtanθ0​−2v02​cos2θ0​gx2​

and

y=2x−9x2,y = 2x - 9x^2,y=2x−9x2,

we get:

(i) Coefficient of xxx

tan⁡θ0=2\tan\theta_0 = 2tanθ0​=2

So we can draw a right triangle with opposite =2=2=2, adjacent =1=1=1, hence hypotenuse =5=\sqrt{5}=5​. Therefore,

sin⁡θ0=25,cos⁡θ0=15\sin\theta_0 = \frac{2}{\sqrt{5}}, \qquad \cos\theta_0 = \frac{1}{\sqrt{5}}sinθ0​=5​2​,cosθ0​=5​1​

Thus,

θ0=tan⁡−1(2)=cos⁡−1(15)=sin⁡−1(25)\theta_0 = \tan^{-1}(2) = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) = \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)θ0​=tan−1(2)=cos−1(5​1​)=sin−1(5​2​)

(ii) Coefficient of x2x^2x2

g2v02cos⁡2θ0=9\frac{g}{2v_0^2\cos^2\theta_0} = 92v02​cos2θ0​g​=9

Substitute g=10g=10g=10 and cos⁡2θ0=(15)2=15\cos^2\theta_0 = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5}cos2θ0​=(5​1​)2=51​:

102v02⋅15=9\frac{10}{2v_0^2\cdot \frac{1}{5}} = 92v02​⋅51​10​=9 102v025=9\frac{10}{\frac{2v_0^2}{5}} = 952v02​​10​=9 502v02=9\frac{50}{2v_0^2} = 92v02​50​=9 25v02=9\frac{25}{v_0^2} = 9v02​25​=9 v02=259v_0^2 = \frac{25}{9}v02​=925​ v0=53 m s−1v_0 = \frac{5}{3}\,\text{m s}^{-1}v0​=35​m s−1
  1. Match with options

We found:

θ0=cos⁡−1(15)\theta_0 = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right)θ0​=cos−1(5​1​)

and

v0=53 m s−1v_0 = \frac{5}{3}\,\text{m s}^{-1}v0​=35​m s−1

This matches Option A.

  1. Check other equivalent form

Since sin⁡θ0=25\sin\theta_0 = \frac{2}{\sqrt{5}}sinθ0​=5​2​, the angle can also be written as

θ0=sin⁡−1(25)\theta_0 = \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)θ0​=sin−1(5​2​)

But among the given options, only A has the correct angle expression together with the correct speed.

Therefore, the correct answer is A.

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