(a) SOLUTION:
[tex]\begin{aligned} y & = \frac{1}{2}gt^2 \\ gt^2 & = 2y \\ t^2 & = \frac{2y}{g} \\ t & = \sqrt{\frac{2y}{g}} \\ & = \sqrt{\frac{\text{2(45 m)}}{\text{9.8 m/s}^2}} \\ & = \boxed{\text{3.03 s}} \end{aligned}[/tex]
Hence, the pen takes 3.03 s to reach the ground.
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(b) SOLUTION:
[tex]\begin{aligned} v^2 & = 2gy \\ v & = \sqrt{2gy} \\ & = \sqrt{2(\text{9.8 m/s}^2)(\text{45 m})} \\ & = \boxed{\text{29.70 m/s}} \end{aligned}[/tex]
Hence, the speed of the pen when it strikes the ground is 29.70 m/s.
[tex]\\[/tex]
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