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Home » 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)
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)
Physics | Physics
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)

Solution: The correct answer is D.

We\,have:\,\,g=\frac{G{{M}_{e}}}{{{R}^{2}}}

Applying\text{ }law\text{ }of\text{ }conservation\text{ }of\text{ }mechanical\text{ }energy:-

\frac{1}{2}m{{u}^{2}}-\frac{G{{M}_{e}}m}{R}=-\frac{G{{M}_{e}}m}{R+h}

\Rightarrow \frac{{{u}^{2}}}{2}-gR=\frac{-g{{R}^{2}}}{R+h}

\frac{g{{R}^{2}}}{R+h}=\frac{2gR-{{u}^{2}}}{2}

R+h=\frac{2g{{R}^{2}}}{2gR-{{u}^{2}}}

h=\frac{{{u}^{2}}R}{2gR-{{u}^{2}}}

i

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