Question

a body is projected vertically upwards with speed of root GM/R. the body will attain a height of

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let The radius of earth be "R".
• Let the Mass of Earth be "M".

$\huge{\bold{\underline{Explanation:-}}}$

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Before Finding, we need to find a Relation,

From Newton's Law of Gravitation.

$\large{\boxed{\tt F = \dfrac{GMm}{R^2}}}$

But This Force is Provided by the Weight of the body,

∴ F = mg.

Substituting in the Formula,

$\large{\tt \leadsto mg = \dfrac{GMm}{T^2}}$

$\large{\tt \leadsto \cancel{m}g = \dfrac{GM\cancel{m}}{R^2}}$

$\large{\boxed{\tt g = \dfrac{GM}{R^2}}}$

The above equation is relation between Acceleration due to gravity(g) and Universal Gravitational Constant(G).

Now, Rearranging the equation we get,

$\large{\tt g = \dfrac{GM}{R^2}}$

$\large{\tt \leadsto gR = \dfrac{GM}{R}}$

$\large{\tt \leadsto gR = \dfrac{GM}{R} \: -----(1)}$

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Now, Given,

The Initial velocity is $\large{\tt u = \sqrt{\dfrac{GM}{R}}}$

$\large{\tt \leadsto u = \sqrt{\dfrac{GM}{R}}}$

Substituting the value from Equation (1).

$\large{\underline{\underline{\tt \leadsto u = \sqrt{gR}}}}$

Now, Applying Third kinematic equation,

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

$\large{\tt \leadsto (0)^2 - (\sqrt{gR})^2 = 2 \times - g \times h}$

• The Final velocity will be zero as the velocity is zero at maximum height.
• Acceleration due to gravity is taken as negative because it is moving against gravity.

$\large{\tt \leadsto 0 - gR = - 2g \times h}$

$\large{\tt \leadsto - gR = - 2g \times h}$

$\large{\tt \leadsto \cancel{-} gR = \cancel{-} 2g \times h}$

$\large{\tt \leadsto gR = 2g \times h}$

$\large{\tt \leadsto h = \dfrac{gR}{2g}}$

$\large{\tt \leadsto h = \dfrac{\cancel{g}R}{2\cancel{g}}}$

$\huge{\boxed{\boxed{\tt h = \dfrac{R}{2}}}}$

So, the Body will reach a height (h) of R/2.

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Answer 2

Answer:h=R

Explanation: