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MHT CET 2026 (12 May, Shift 2) · Escape Velocity

Escape Velocity question from MHT CET 2026 (12 May, Shift 2)

Q5MHT CET 2026 · 12 MAY · SHIFT 2Medium

Escape Velocity

A body is projected vertically upwards from earth's surface. If its kinetic energy of projection is equal to half of its minimum value required to escape from the gravitational influence then the height up to which it rises is (RR = radius of earth)

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Answer: (c)

  1. For v=Kvev = Kv_e: h=RK21−K2h = \frac{RK^2}{1-K^2}.
  2. KE is half of escape KE, so K2=12K^2 = \frac{1}{2}.
  3. h=R⋅121−12=Rh = \frac{R\cdot\frac{1}{2}}{1-\frac{1}{2}} = R.
  4. So the answer is (c) RR.
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