30. (a) For the ball, we have $u=45 m{s}^{-1},g=-10 m{s}^{-2}$

Now using, $v^2-u^2=2 g h$, we have

$v^2=(45{)}^2-20 h$

$\Rightarrow v=\sqrt{2025-20h}$

At $v=0,h=\frac{2025}{20}=101 m$

at $h=0,v=45 m{s}^{-1}$

As velocity decreases-with height, slope of $v-h$ graph must be negative at all points. Hence, correct graph is (a).