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15/09/2021 1:00 pm
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A particle of mass 'm' is projected with a velocity v = kVe (k < 1) from the surface of the earth.
(ve = escape velocity)
The maximum height above the surface reached by the particle is:
(1) \(R(\frac{k}{1-k})^2\)
(2) \(R(\frac{k}{1+k})^2\)
(3) \(\frac{R^2k}{1+k}\)
(4) \(\frac{Rk^2}{1-k^2}\)
1 Answer
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15/09/2021 1:05 pm
Correct answer: (4) \(\frac{Rk^2}{1-k^2}\)
Explanation:
h = \(\frac{R}{\frac{2gR}{V^2}-1}\)
= \(\frac{R}{\frac{V^2_e}{K^2V_e^2}-1}\)
= \(\frac{Rk^2}{1-k^2}\)