A body is projected from the ground vertically upwards with a speed of 4 km/s. How high will it rise ? Neglect air resistance. (Radius of the earth = 6400 km, Acceleration due to gravity $= 9.8 m / s^{2}$

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Solution

Decrease in kinetic energy $=$ increase in gravitational potential energy of particle
$\frac{1}{2} m v^{2} = \frac{m g h}{1 + \frac{h}{R}}$
$∴ h = \frac{v^{2}}{2 g - \frac{v^{2}}{R}}$
$= \frac{{(4.0 \times 10^{3})}^{2}}{2 \times 9.8 - \frac{{(4.0 \times 10^{3})}^{2}}{6.4 \times 10^{6}}}$
$= 9.35 \times 10^{5} m$
we get, $h = 935 k m$

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A body is projected from the ground vertically upwards with a speed of 4 km/s. How high will it rise ? Neglect air resistance. (Radius of the earth = 6400 km, Acceleration due to gravity =9.8m/s2

Decrease in kinetic energy=increase in gravitational potential energy of particle 12mv2=mgh1+hR∴h=v22g−v2R=(4.0×103)22×9.8−(4.0×103)26.4×106=9.35×105mwe get, h=935km

A body is projected from the ground vertically upwards with a speed of 4 km/s. How high will it rise ? Neglect air resistance. (Radius of the earth = 6400 km, Acceleration due to gravity $= 9.8 m / s^{2}$