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A body fell down from rest and is falling down with an acceleration 10 m/s2. When it was hitting the ground, its displacement was 20 m. The velocity with which it hits the ground is:

A5 m/s

B10 m/s

C15 m/s,

D20 m/s

Answer:

D. 20 m/s

Read Explanation:

Understanding Motion Under Constant Acceleration

  • This problem involves a body undergoing uniformly accelerated motion, specifically falling under gravity.
  • The key to solving such problems is identifying the known quantities and selecting the appropriate equation of motion.

Given Parameters:

  • Initial velocity (u) = 0 m/s (since the body fell from rest).
  • Acceleration (a) = 10 m/s² (given). This is often approximated as the acceleration due to gravity (g) for simplification in competitive exams.
  • Displacement (s) = 20 m.

Applying the Equation of Motion:

  • The most suitable kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s) is: v² = u² + 2as.
  • Substituting the given values into the equation:
    • v² = (0)² + 2 × 10 m/s² × 20 m
    • v² = 0 + 400 m²/s²
    • v² = 400 m²/s²
    • v = √400 m²/s²
    • v = 20 m/s
  • Therefore, the velocity with which the body hits the ground is 20 m/s.

Key Facts for Competitive Exams:

  • There are three primary equations of motion for uniformly accelerated linear motion:
    1. v = u + at (Velocity-time relation)
    2. s = ut + ½at² (Position-time relation)
    3. v² = u² + 2as (Position-velocity relation)
  • For objects falling freely under gravity, the acceleration 'a' is replaced by 'g' (acceleration due to gravity), which is approximately 9.8 m/s² or often taken as 10 m/s² for calculation simplicity in exams.
  • When a body falls from rest, its initial velocity (u) is always 0.
  • The direction of acceleration due to gravity (g) is always downwards, towards the center of the Earth.
  • Air resistance is generally neglected in these types of ideal physics problems unless specified.

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