Question

A body is thrown from the surface of the earth with velocity 'u' m/s. The maximum height in m above the surface of the earth upto which it will reach is ______.

(R = radius of earth, g = acceleration due to gravity)

Options

• `("u"^2"R")/(2"gR"-"u"^2)`

• `(2"u"^2"R")/("gR"-"u"^2)`

• `("u"^2"R"^2)/(2"gR"^2-"u"^2)`

• `("u"^2"R")/("gR"-"u"^2)`

Answer 1

A body is thrown from the surface of the earth with velocity 'u' m/s. The maximum height in m above the surface of the earth upto which it will reach is `underlinebb(("u"^2"R")/(2"gR"-"u"^2))`.

(R = radius of earth, g = acceleration due to gravity)

Explanation:

Let 'h' be the height of the body above the surface of the earth.

`"GMm"/"R"+1/2"mu"^2=0+"GMm"/("R"+"h")`

`"GM"/("R"+"h")="Gm"/"R"-
"u"^2/2`

`"GM"/(("R"+"h"))=(2"Gm"-"Ru"^2)/(2"R")`

`("R"+"h")/"GM"=(2"R")/(2"GM"-"Ru"^2)`

h = `(2"GMR")/(2"GM"-"Ru"^2)-"R"`

= `(2"GMR"-2"GMR"+"R"^2"u"^2)/(2"GM"-"Ru"^2)`

= `("R"^2"u"^2)/(2"GM"-"Ru"^2)`

= `"Ru"^2/(2"gR"-"u"^2)`