A ball is projected at $72km.h^{-1}$ speed from the top of a building. The building is $72$ metres tall. The distance travelled in metres before it reaches the ground is as follows,

$s=u_{0}t+\frac{1}{2}g{t}^2$

where: $u_0=$ initial velocity $(m*s^{-1})$

$g=$ acceleration due $to$ gravity $(9\mathrm{.}8(m)*s^{-2})$

$t=$ time $(s)$

a$)$

What is the speed of the ball of in $m*s^{-1}?$

$$ m$.s^{-1}$