A body is projected vertically upwards from the ground with a velocity '' and it gains a maximum height 'H'm. At a point 'x'm height (x < H) the ratio f its potential to kinetic energy is $1 : 2$ . Now if the body is projected upwards with the velocity ' $3$ u' from the ground, find the potential to kinetic energy at a point whose height is $3$ 'x'm.

4 : 1

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1 : 8

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18 : 1

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16 : 1

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Solution

The correct option is B $1 : 8$
Initial Energy $= K E_{i} = \frac{1}{2} m u^{2}$

At Point x $K E_{x} = \frac{1}{2} m u^{2} - m g x \frac{m g x}{\frac{1}{2} m u^{2} - m g x} = \frac{1}{2} 2 m g x = \frac{1}{2} m u^{2} - m g x 3 m g x = \frac{1}{2} m u^{2} 3 x = \frac{u^{2}}{2 g}$

Now if $u^{1} = 3 u x^{1} = 3 x$

$K E_{3 x} = \frac{1}{2} m {u^{1}}^{2} - m {g x}^{1} \Rightarrow \frac{1}{2} m 9 u^{2} - 3 m g x P E_{3 x} = m g$

$f^{1} = \frac{3 m g x}{\frac{9}{2} m u^{2} - 3 m g x} \Rightarrow \frac{3 x}{\frac{9 u^{2}}{2 g} - 3 x} \Rightarrow \frac{3 x}{27 x - 3 x} = \frac{3 x}{24 x} = 1 : 8$

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The correct option is B 1:8Initial Energy =KEi=12mu2 At Point x KEx=12mu2−mgxmgx12mu2−mgx=122mgx=12mu2−mgx3mgx=12mu23x=u22g Now if u1=3ux1=3x KE3x=12mu12−mgx1⇒12m9u2−3mgxPE3x=mg f1=3mgx92mu2−3mgx⇒3x9u22g−3x⇒3x27x−3x=3x24x=1:8