A stone is thrown vertically upwards with a velocity of 40 m/s. At what height will its kinetic energy and potential energy be equal?

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Solution

$T E = K E + P E$
$f o r e x a m p l e$
$10 J = 5 J + 5 J; a s w e c a n s e e, K E = P E o n l y w h e n b o t h e n e r g i e s a r e e q u a l a t o n e p o i n t d u r i n g t h e i r j o u r n e y .$
Alternative solution.
TE=KE+PE
$\frac{1}{2} m v^{2} = \frac{1}{2} m v^{' 2} + m g h^{'}$
$b u t h e i g h t a t w h i c h K E a n d P E a r e e q u a l$

$\frac{1}{2} m v^{' 2} = m g h^{'}$

$\frac{1}{2} m v^{2} = m g h^{'} + m g h^{'}$
$h^{'} = \frac{v^{2}}{4 g}$

$h^{'} = 40 m$

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