Maximum height reached by an object projected | vedclass

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(C) The escape velocity from the surface of the Earth is given by $v_e = \sqrt{2gR}$.Given that the initial speed of the object is $v = 50\% 	ext{ of

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$\frac{R}{3}$

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(C) The escape velocity from the surface of the Earth is given by $v_e = \sqrt{2gR}$.Given that the initial speed of the object is $v = 50\% 	ext{ of } v_e = \frac{v_e}{2} = \frac{\sqrt{2gR}}{2}$.Using the principle of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$:$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$.Substituting $v = \frac{\sqrt{2gR}}{2}$ and $GM = gR^2$:$\frac{1}{2}m \left(\frac{2gR}{4}ight) - \frac{mgR^2}{R} = - \frac{mgR^2}{R+h}$.$\frac{mgR}{4} - mgR = - \frac{mgR^2}{R+h}$.$-\frac{3}{4}mgR = - \frac{mgR^2}{R+h}$.$\frac{3}{4} = \frac{R}{R+h}$.$3(R+h) = 4R \Rightarrow 3R + 3h = 4R \Rightarrow 3h = R \Rightarrow h = \frac{R}{3}$.

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 the earth's surface is - ($R =$ Radius of Earth)

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