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Motion in A Plane question

2022 · 26 Jul · Shift 2 · Q40
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Motion in A Plane question

2022 · 26 Jul · Shift 2 · Q40

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
Two projectiles are thrown with same initial velocity making an angle of 45∘45^{\circ}45∘ and 30∘30^{\circ}30∘ with the horizontal respectively. The ratio of their respective ranges will be :
  1. A
    1:21: \sqrt{2}1:2​
  2. B
    2:1\sqrt{2}: 12​:1
  3. C
    2:32: \sqrt{3}2:3​
  4. D
    3:2\sqrt{3}: 23​:2
View written solutionFree

Correct answer: C

  1. Use the formula for horizontal range of a projectile

    For a projectile projected with speed uuu at angle θ\thetaθ, the horizontal range is R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}R=gu2sin2θ​

  2. First projectile: angle 45∘45^\circ45∘

    R1=u2sin⁡(2⋅45∘)g=u2sin⁡90∘g=u2gR_1 = \frac{u^2 \sin(2\cdot 45^\circ)}{g} = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g}R1​=gu2sin(2⋅45∘)​=gu2sin90∘​=gu2​

  3. Second projectile: angle 30∘30^\circ30∘

    R2=u2sin⁡(2⋅30∘)g=u2sin⁡60∘g=u2g⋅32R_2 = \frac{u^2 \sin(2\cdot 30^\circ)}{g} = \frac{u^2 \sin 60^\circ}{g} = \frac{u^2}{g}\cdot \frac{\sqrt{3}}{2}R2​=gu2sin(2⋅30∘)​=gu2sin60∘​=gu2​⋅23​​

  4. Find the ratio

    R1:R2=u2g:u2g⋅32R_1 : R_2 = \frac{u^2}{g} : \frac{u^2}{g}\cdot \frac{\sqrt{3}}{2}R1​:R2​=gu2​:gu2​⋅23​​

    Cancel the common factor u2g\frac{u^2}{g}gu2​:

    R1:R2=1:32=23:1=2:3R_1 : R_2 = 1 : \frac{\sqrt{3}}{2} = \frac{2}{\sqrt{3}} : 1 = 2 : \sqrt{3}R1​:R2​=1:23​​=3​2​:1=2:3​

  5. Match with the options

    2:32 : \sqrt{3}2:3​ corresponds to Option C.

Final Answer: 2:3\boxed{2: \sqrt{3}}2:3​​

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