Answer
B. $2t$ and $4Y$
Work Step by Step
Case first. When launching velocity is $V_°
$
When the ball reach the top it's launching velocity becomes zero.
According to first equation of motio $v$ = $u$ $+$ $at$
$0$ $=$ $V$$_°$ $+$ ($-$$gt$)
$V$$_°$ $=$ $gt$
$t$ $=$ $V$$_°$$/$$g$
According to third equation of motion
$v^2$ $=$ $u^2$ $+$ $2as$
$0$ $=$ $V$$_°$$^2$ $+$ $2$($-g$)$Y$
$Y$ $=$$V$$_°$$^2$$/$$2g$
In case second when launching velocity is $2$$V$$_°$$^2$
Similarly
$t'$ $=$ $2$$V$$_°$$^2$$/$$g$ $= $ $2t$
And
$Y'$ $=$ $($$2V$$_°$$)$$^2$$/$$2g$ $=$ $4V$$_°$$^2$$/$$2g$ $=$ $4Y$
