Question

A body projected up with a velocity "u" reaches height “h". To reach double the height, it must be projected up with a velocity of?​

Answer 1

By third equation of motion

$v^{2} -u^{2} =2as\\0-u^{2} =2(-g)h\\u^{2} =2gh ---1\\\\For \ height \ 2h\\v^{2}- u^{'2} =2(-g)2h\\u^{'2}=4gh---2\\hence \\from \ 1\ and\ 2\\u^{'2} =2u^{2} \\u'=\sqrt{2} u$