AP EAMCET · PHYSICS · Gravitation
Maximum height reached by a rocket fired with a speed equal to \(50 \%\) of the escape speed from the surface of the earth is ( \(R\) - Radius of the earth)
- A \(\frac{R}{2}\)
- B \(\frac{16 R}{9}\)
- C \(\frac{\mathrm{R}}{3}\)
- D \(\frac{\mathrm{R}}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{R}}{3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{v}_{\mathrm{i}}=\frac{1}{2} \mathrm{v}_{\mathrm{e}}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}, \mathrm{U}_{\mathrm{f}}=0\) Applying conservation of mechanical energy, \(K_i+V_i=K_f+U_f\)…
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