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NEET PHYSICSMOTION IN A STRAIGHT LINEEasy

Question

When a ball is thrown up vertically with velocity v0v_0, it reaches a maximum height of 'hh'. If one wishes to triple the maximum height then the ball should be thrown with velocity:

A

3v0\sqrt{3}v_0

B

3v03v_0

C

9v09v_0

D

(3/2)v0(3/2)v_0

Step-by-Step Solution

  1. Identify the Relationship: For a body thrown vertically upwards with initial velocity uu, the maximum height HH reached is given by the kinematic equation v2=u2+2asv^2 = u^2 + 2as. At maximum height, final velocity v=0v=0 and acceleration a=ga = -g. 0=u22gH    u=2gH0 = u^2 - 2gH \implies u = \sqrt{2gH} This shows that the initial velocity is directly proportional to the square root of the maximum height (uHu \propto \sqrt{H}) .
  2. Apply the Condition:
  • Case 1: Initial velocity =v0= v_0, Max height =h= h. So, v0hv_0 \propto \sqrt{h}.
  • Case 2: New max height h=3hh' = 3h. Let the new velocity be vv'. v3h=3×hv' \propto \sqrt{3h} = \sqrt{3} \times \sqrt{h}
  1. Determine New Velocity: Substituting v0v_0 for h\sqrt{h} (proportionality constant cancels out): v=3v0v' = \sqrt{3} v_0

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 LINEthrownverticallyvelocityreachesmaximum

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