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A body of mass 2kg projected with velocityV = \frac{V_{e}}{2} from the earth, if the maximum height attains by the body is h then will be (here Ve is the escape velocity).

Option: 1

\frac{1}{3}R


Option: 2

\frac{2}{3}R


Option: 3

\frac{1}{2}R


Option: 4

\frac{4}{3}R


Answers (1)

best_answer

As we learn

Maximum height attained by the body -

h=R\left [ \frac{V^{2}}{V_{e}^{2}-V^{2}} \right ]

V_{e}\rightarrow escape velocity

V\rightarrow Projection velocity of body

R\rightarrow Radius of planet

- wherein

By conservation of energy

Total Energy at surface = Total Energy at height h 

 

 maximum height attains by the body 

h = (\frac{v^{2}}{v_{e}^{2}-v^{2}})R

v = \frac{v_{e}}{2}

h= \frac{\frac{v_{e}^{2}}{4}}{v_{e}^{2}-\frac{v_{e}^{2}}{4}}R

h= \frac{R}{3}

 

Posted by

Irshad Anwar

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