Solution to Test Paper 2

& 44.1=\frac{1}{2}\times 9.8{{t}^{2}} \\ & t=\sqrt{\frac{44.1}{4.9}}=3\text{ s} \\ \end{align}

OR

\displaystyle \begin{align} & s=\frac{1}{2}\times 9.8\times {{3}^{2}}=44.1 \\ & \text{AG} \\ & \therefore \text{Hits ground after 3 seconds} \\ \end{align}