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A particle is released from a height S. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively |
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Answer» `S/4, (3gS)/(2)` `v^2 = 0 + 2g(S - x)""..(i)` or `v^2 = 2g(S - x)` potential energy at `B = mgx` `:. ` Kinetic energy = `3 xx "potential energy"` `:. 1/2 m xx 2 g (S - x) = 3 xx mgx` or `S - x = 3x " or " S = 4x " or " x = S/4` Putting this value of x in EQN. (i), we get `v^2= 2g (S - S/4) = (2g xx 3S)/(4) = (3gS)/(2) "or" v= sqrt((3gS)/(2))` Hence, `x = S/4` and `v = sqrt((3gS)/(2))`
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