If object is projected upwards from surface of earth with speed equal to Ve/3 (Ve=escape speed). find the maximum height reached by the object from surface of earth?
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Answers
Answer:
We know that the escape velocity of Earth = Ve = √[2gR]
g = gravity due to Earth
1/2 m (Ve)² - G Mm/R = 0 for the object to escape from Earth.
(or, KE + PE = 0)
When the body is projected from Earth's surface:
Given u = initial velocity = Ve /2 = √(gR/2)
Initial KE = 1/2 m u² = m g R/4 = G M m /(4R)
Initial PE = - GMm/R
Total initial energy = - 3 GMm /(4R)
When the body reaches a height h above surface of Earth and stops:
v = 0. Final KE = 0.
Final PE = - G M m/ (R+h)
From conservation of Energy:
- GMm/(R+h) = - 3Gm/ (4R)
3 (R+h) = 4 R
h = R/3
Answer is R/3.
diagram of mitotic metaphase i hope this might help u .
sorry this is not my own drawing i took this from go ogle
