A particle of mass ' m ' is projected with a velocity v = kVe(k &lt; 1) from the surface of the earth. (Ve = escape velocity) The maximum height above the surface reached by the particle is: &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

R(k²/(1-k²))

R((1-k²)/k²)

R²k/(1+k)

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Overview Physics Gravitation Physics Class 11th

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:

Option 1 -

R(k²/(1-k²))

Option 2 -

R((1-k²)/k²)

Option 3 -

R²k/(1+k)

Option 4 -

Rk²/(1-k²)

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 weeks ago

Correct Option - 4

Detailed Solution:

h = R / [ (2gR/V? ²) - 1] = R / [ (V²/K²V? ²) ] = KK² / (1 - K²)

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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:

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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:

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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: