physics
posted by Annointed Jesu on .
an object is thrown upward with an initial velocity 'u' m/s. derive an expression for the height 'h' at which its kinetic energy becomes equal to its potential energy.

h=(u² v²)/2g
v²=u² 2gh,
KE=m•v²/2= m(u² 2gh)/2,
KE=PE =>
m(u² 2gh)/2 =mgh,
(u² 2gh)/2= gh,
u²=4gh,
h= u²/4g.