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  • A projectile is launched from the surface of the earth with a very high speed v at an angle with vertical. What is its velocity when it is at the farthest distance from the earth surface, given tha

A projectile is launched from the surface of the earth with a very high speed v at an angle with vertical. What is its velocity when it is at the farthest distance from the earth surface, given that the maximum height reached by the projectile is equal to the height reached when it is launched perpendicular to earth with a velocity =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}

Option: 1

\mathrm{\frac{v \cos \theta}{2}}


Option: 2

\mathrm{\frac{v \sin \theta}{2}}


Option: 3

\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}


Option: 4

\sqrt{\frac{\mathrm{GM}}{3 \mathrm{R}}}


Answers (1)

best_answer

The maximum height reached by the projectile is given by

\mathrm{\frac{1}{2} m v^2-\frac{G M m}{R}=-\frac{G M m}{R+h}}

\mathrm{\text { or,. } \frac{1}{2} m \frac{G M}{R}-\frac{G M m}{R}=-\frac{G M m}{R+h}=-\frac{G M m}{2 R}}

\mathrm{\therefore \quad \mathrm{h}=\mathrm{R}}

Applying conservation of angular momentum

\mathrm{\begin{aligned} &m u^{\prime}(R+h)=m v \sin \theta R\\ &\therefore \quad u^{\prime}(2 R)=v \sin \theta R\\ &\therefore \quad u^{\prime}=\frac{v \sin \theta}{2} \end{aligned}}

Hence, (B) is correct

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