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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=u2−2gH  ⟹  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 (u∝Hu \propto \sqrt{H}) .
  2. Apply the Condition:
  • Case 1: Initial velocity =v0= v_0, Max height =h= h. So, v0∝hv_0 \propto \sqrt{h}.
  • Case 2: New max height h′=3hh' = 3h. Let the new velocity be v′v'. v′∝3h=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
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