Question

A body is thrown from the surface of the earth velocity v `"m"/"s"`. The maximum height above the earth's surface upto which it will reach is ____________.

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

विकल्प

• `("vR")/(2"gR" - "v")`

• `(2"gR")/("v"^2("R" - 1))`

• `("vR"^2)/("gR" - "v")`

• `("v"^2"R")/(2"gR" - "v"^2)`

Answer 1

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

Explanation:

At maximum height, the velocity of body becomes zero. Let maximum height be h.

Applying the principle of conservation of energy, we get

`1/2 "mv"^2 - ("GMm")/"R" = 1/2"m"(0)^2 - ("GMm")/(("R" + "h"))` .......[where, m = mass of body and M = mass of earth.]

`1/2 "mv"^2 = "GMm" (1/"R" - 1/("R" + "h"))`

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

`"h"/("R" + "h") = ("v"^2"R")/(2"GM") = ("v"^2"R")/(2"gR"^2)` ..........(∵ GM = gR²)

`("R" + "h")/"h" = (2"gR")/"v"^2`

`"R"/"h" + 1 = (2"gR")/"v"^2`

`"R"/"h" = (2"gR")/"v"^2 - 1 = (2"gR" - "v"^2)/"v"^2`

or h = `("v"^2"R")/(2"gR" - "v"^2)`