InterviewSolution
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InterviewSolution
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A projectile is fired from the surface of earth of radius `R` with a velocity `k upsilon_(e)` (where `upsilon_(e)` is the escape velocity from surface of earth and `k lt 1)`. Neglecting air resistance, the maximum height of rise from centre of earth isA. `(I-k^(2))/R`B. `R/(1-k^(2))`C. `R(1-k)^(2)`D. `R/(1+k^(2))` |
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Answer» Correct Answer - B If a body is projected from the surface of the earth with a velocity v and reaches a height h, then according to law of conservation of energy `1/2mv^(2)=(mgh)/(1+h//R)` Here `v=kv_(e)=ksqrt(2gR)` Therefore, `1/2mk^(23)gR=(mg(r-R))/(1+(r-R)/R)` `k^(2)R[1+(r-R)/R]=r-R` or `k^(2)r=r-R` or `r=(R/(1-k^(2)))` |
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