Question

A body is projected vertically from earth's surface of radius R with velocity equal to half the escape velocity. The maximum height reached by the body is

• A. \(\frac{R}{3}\)
• B. \(R\)
• C. \(\frac{R}{2}\)
• D. \(\frac{R}{4}\)

Hint:

Use energy conservation with the gravitational potential energy and the value of v squared from half the escape speed.

Answer 1

The Correct Option is A

Step 1: Use the general formula.
For launch speed $v$, the maximum height is $h = \dfrac{R}{\dfrac{v_e^2}{v^2} - 1}$.

Step 2: Put the numbers.
With $v = v_e/2$: $v_e^2/v^2 = 4$, so $h = \dfrac{R}{4 - 1} = \dfrac{R}{3}$.

Step 3: Check with energy.
At $r = 4R/3$: potential energy is $-\dfrac{3GMm}{4R}$. At the surface, $KE + PE = \dfrac{GMm}{4R} - \dfrac{GMm}{R} = -\dfrac{3GMm}{4R}$, which agrees.

Final Answer:
Option (A). \[ \boxed{\frac{R}{3}} \]