If a ball is thrown at a velocity of 45m / s in vertical upward...

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

Now using, ${v}^{2} - {u}^{2} = 2gh$, we have
${v}^{2} = ( 45{)}^{2} - 20h$

$⇒ v = \sqrt{2025 - 20h}$
At $v = 0, h = \dfrac{2025}{20} = 101m$

at $h = 0, v = 45m{s}^{- 1}$
As velocity decreases-with height, slope of $v - h$ graph must be negative at all points. Hence, correct graph is (a).