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NEET PHYSICSMOTION IN A STRAIGHT LINEEasy

Question

A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 ms⁻², the velocity with which it hits the ground is:

A

20 m/s

B

40 m/s

C

5 m/s

D

10 m/s

Step-by-Step Solution

  1. Identify Given Values: Initial velocity (v0v_0) = 0 (dropped from rest). Displacement (yy) = 20 m (taking downward direction as positive).
  • Acceleration (aa) = gg = 10 m/s².
  1. Select Kinematic Equation: Use the equation connecting velocity, displacement, and acceleration: v2=v02+2ayv^2 = v_0^2 + 2ay
  2. Calculation: v2=02+2(10)(20)v^2 = 0^2 + 2(10)(20) v2=400v^2 = 400 v=400=20 m/sv = \sqrt{400} = 20 \text{ m/s}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEstandingheightassumingvelocityground

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