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7.Gravitation
normal
Maximum height reached by an object projected perpendicular to the surface of the earth with a speed equal to $50\%$ of the escape velocity from earth surface is - ( $R =$ Radius of Earth)
A
$\frac {R}{2}$
B
$\frac {16R}{9}$
C
$\frac {R}{3}$
D
$\frac {R}{8}$
Solution
$\frac{1}{2} \mathrm{mV}^{2}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}$
$\frac{1}{2} \mathrm{m}\left(\frac{\mathrm{v}_{\mathrm{e}}}{2}\right)^{2}=\frac{\mathrm{mgh} \mathrm{R}}{\mathrm{R}+\mathrm{h}}$
$\frac{1}{8} \times 2 \mathrm{gR}=\frac{\mathrm{ghR}}{\mathrm{R}+\mathrm{h}}$
$\frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}=\frac{1}{4} \Rightarrow \mathrm{h}=\frac{\mathrm{R}}{3}$
Standard 11
Physics
