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Motion in A Plane question

2023 · 1 Feb · Shift 1 · Q53
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  5. /2023 · 1 Feb · Shift 1 · Q53

Motion in A Plane question

2023 · 1 Feb · Shift 1 · Q53

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A child stands on the edge of the cliff 10 m10 \mathrm{~m}10 m above the ground and throws a stone horizontally with an initial speed of 5 ms−15 \mathrm{~ms}^{-1}5 ms−1. Neglecting the air resistance, the speed with which the stone hits the ground will be ms−1\mathrm{ms}^{-1}ms−1(given, g=10 ms−2g=10 \mathrm{~ms}^{-2}g=10 ms−2 ).
  1. A
    20
  2. B
    25
  3. C
    30
  4. D
    15
View written solutionFree

Correct answer: D

  1. Resolve motion into horizontal and vertical components

The stone is thrown horizontally with speed ux=5 m s−1u_x = 5\,\text{m s}^{-1}ux​=5m s−1 and initial vertical speed uy=0.u_y = 0.uy​=0.

  1. Find the vertical speed just before hitting the ground

The stone falls through a vertical height h=10 m.h = 10\,\text{m}.h=10m.

Using vy2=uy2+2gh,v_y^2 = u_y^2 + 2gh,vy2​=uy2​+2gh, we get vy2=0+2(10)(10)=200.v_y^2 = 0 + 2(10)(10) = 200.vy2​=0+2(10)(10)=200. So, vy=200=102 m s−1.v_y = \sqrt{200} = 10\sqrt{2}\,\text{m s}^{-1}.vy​=200​=102​m s−1.

  1. Horizontal speed remains constant

Since air resistance is neglected, vx=5 m s−1.v_x = 5\,\text{m s}^{-1}.vx​=5m s−1.

  1. Resultant speed at impact

The speed with which the stone hits the ground is v=vx2+vy2.v = \sqrt{v_x^2 + v_y^2}.v=vx2​+vy2​​.

Thus, v=52+(102)2=25+200=225=15 m s−1.v = \sqrt{5^2 + (10\sqrt{2})^2} = \sqrt{25 + 200} = \sqrt{225} = 15\,\text{m s}^{-1}.v=52+(102​)2​=25+200​=225​=15m s−1.

  1. Option check
  • A: 202020 ❌
  • B: 252525 ❌
  • C: 303030 ❌
  • D: 151515 ✅

Therefore, the correct answer is D.

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