Question

A particle of mass \(m\) is projected with a velocity \(v=KV_{e}\) (\(K\lt 1\)) from the surface of the earth. The maximum height above the surface reached by the particle is (\(V_{e}\) = escape velocity, \(R\) is radius of earth)

• A. \(R\left(\frac{K}{1+K}\right)^{2}\)
• B. \(\frac{R^{2}K}{1+K}\)
• C. \(\frac{RK^{2}}{1-K^{2}}\)
• D. \(R\left(\frac{K}{1-K}\right)^{2}\)

Hint:

Apply conservation of energy between the surface and the highest point using the escape velocity expression.

Answer 1

The Correct Option is C

Step 1: Formula first
For launch speed $v$, $h=\dfrac{R}{\dfrac{V_e^2}{v^2}-1}$, derived from energy conservation.

Step 2: Substitute
With $v=KV_e$, $\dfrac{V_e^2}{v^2}=\dfrac{1}{K^2}$.

Step 3: Simplify
$h=\dfrac{R}{\dfrac{1}{K^2}-1}=\dfrac{RK^2}{1-K^2}$, option (C).

Final Answer:
The height reached is R K squared over 1 minus K squared. \[ \boxed{\text{(C) }\dfrac{RK^{2}}{1-K^{2}}} \]