Answer
(a) Height of the building is $17.68 \space m$.
(b) Brick strikes the ground with speed $18.62 \space m/s$.
(c) Acceleration, velocity and position vs time graphs will be as shown in figure-
Work Step by Step
(a) From $s=ut+\frac{1}{2}at^2$ (taken downward direction to be positive)
$\Rightarrow h=0\times t+\frac{1}{2}gt^2=0+\frac{1}{2}\times9.8\times1.90^2\approx17.68 \space m$
So height of the building is $17.68 \space m$.
(b) From $v=u+at$
$\Rightarrow v=0+gt$
$\Rightarrow v=9.8\times 1.90=18.62 \space m/s$
So, brick strikes the ground with speed $18.62 \space m/s$.
(c) Taken $y=0$ at the ground and upward direction as positive $y$, acceleration, velocity and position vs time graphs will be as shown in figure-
