ExamBro
ExamBro
AP EAMCET · PHYSICS · Gravitation

An object is thrown directly away from the surface of the earth with an initial speed \(v\). The object reaches upto a height of \(\frac{4}{5} R_E\) from earth's surface, where \(R_E\) is radius of the earth. If the escape velocity of the object is \(v_E\) then the value of \(\frac{v}{v_E}\) is

  1. A \(4 / 3\)
  2. B \(3 / 4\)
  3. C \(2 / 3\)
  4. D \(4 / 5\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2 / 3\)

Step-by-step Solution

Detailed explanation

We know that, maximum height attained by a projectile projected with velocity \(v\). \[ h=\frac{v^2 R_E}{2 g R_E-v^2} \] But given, \(h=\frac{4}{5} R_E\)…