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Motion in A Plane question

2012 · Q154
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Motion in A Plane question

2012 · Q154

NEETPhysicsMotion in A PlaneMCQ+4 / −1
The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
  1. A
    θ\thetaθ = tan−-−1(14)\left( {{1 \over 4}} \right)(41​)
  2. B
    θ\thetaθ = tan−-−1(4)
  3. C
    θ\thetaθ = tan−-−1(2)
  4. D
    θ\thetaθ = 45o
View written solutionFree

Correct answer: B

Horizontal range

R=u2sin⁡2θgR = {{{u^2}\sin 2\theta } \over g}R=gu2sin2θ​     ......(1)

Maximum height

H=u2sin⁡2θ2gH = {{{u^2}{{\sin }^2}\theta } \over {2g}}H=2gu2sin2θ​       .....(2)

According to the problem R = H

u2sin⁡2θg{{{u^2}\sin 2\theta } \over g}gu2sin2θ​ = u2sin⁡2θ2g{{{u^2}{{\sin }^2}\theta } \over {2g}}2gu2sin2θ​

⇒2sin⁡θcos⁡θ=sin⁡2θ2 \Rightarrow 2\sin \theta \cos \theta = {{{{\sin }^2}\theta } \over 2}⇒2sinθcosθ=2sin2θ​

2cos⁡θ=sin⁡θ22\cos \theta = {{\sin \theta } \over 2}2cosθ=2sinθ​

⇒cot⁡θ=14 \Rightarrow \cot \theta = {1 \over 4}⇒cotθ=41​

⇒tan⁡θ=4 \Rightarrow \tan \theta = 4⇒tanθ=4

⇒θ=tan⁡−1(4) \Rightarrow \theta = {\tan ^{ - 1}}(4)⇒θ=tan−1(4)

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