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Motion in A Plane question

2023 · 11 Apr · Shift 2 · Q54
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  5. /2023 · 11 Apr · Shift 2 · Q54

Motion in A Plane question

2023 · 11 Apr · Shift 2 · Q54

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A projectile is projected at 30∘30^{\circ}30∘ from horizontal with initial velocity 40 ms−140 \mathrm{~ms}^{-1}40 ms−1. The velocity of the projectile at t=2 s\mathrm{t}=2 \mathrm{~s}t=2 s from the start will be : (Given g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^{2}g=10 m/s2 )
  1. A
    203 ms−120 \sqrt{3} \mathrm{~ms}^{-1}203​ ms−1
  2. B
    Zero
  3. C
    20 ms−120 \mathrm{~ms}^{-1}20 ms−1
  4. D
    403 ms−140 \sqrt{3} \mathrm{~ms}^{-1}403​ ms−1
View written solutionFree

Correct answer: A

  1. Resolve the initial velocity into components

Given:

  • Initial speed u=40 m s−1u = 40\,\text{m s}^{-1}u=40m s−1
  • Angle of projection θ=30∘\theta = 30^\circθ=30∘
  • Acceleration due to gravity g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2

Horizontal component: ux=ucos⁡30∘=40⋅32=203 m s−1u_x = u\cos 30^\circ = 40\cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}\,\text{m s}^{-1}ux​=ucos30∘=40⋅23​​=203​m s−1

Vertical component: uy=usin⁡30∘=40⋅12=20 m s−1u_y = u\sin 30^\circ = 40\cdot \frac{1}{2} = 20\,\text{m s}^{-1}uy​=usin30∘=40⋅21​=20m s−1

  1. Find velocity components after t=2 t=2\,t=2s

  • Horizontal velocity remains constant: vx=203 m s−1v_x = 20\sqrt{3}\,\text{m s}^{-1}vx​=203​m s−1

  • Vertical velocity changes due to gravity: vy=uy−gt=20−10×2=0v_y = u_y - gt = 20 - 10\times 2 = 0vy​=uy​−gt=20−10×2=0

  1. Find the resultant velocity at t=2 t=2\,t=2s

Speed is: v=vx2+vy2v = \sqrt{v_x^2 + v_y^2}v=vx2​+vy2​​

Substitute values: v=(203)2+02=203 m s−1v = \sqrt{(20\sqrt{3})^2 + 0^2} = 20\sqrt{3}\,\text{m s}^{-1}v=(203​)2+02​=203​m s−1

  1. Match with the options
  • A: 203 m s−120\sqrt{3}\,\text{m s}^{-1}203​m s−1 ✅
  • B: Zero ❌
  • C: 20 m s−120\,\text{m s}^{-1}20m s−1 ❌
  • D: 403 m s−140\sqrt{3}\,\text{m s}^{-1}403​m s−1 ❌

Therefore, the correct answer is: 203 m s−1\boxed{20\sqrt{3}\,\text{m s}^{-1}}203​m s−1​

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