In order to calculate the time needed for the firework to reach 201 feet, let's use h(t) = 201 and v0 = 130 in the equation. Then, we put it in the standard form and calculate the zeros using the quadratic formula:
![\begin{gathered} h(t)=-16t^2+v_0t \\ 201=-16t^2+130t \\ -16t^2+130t-201=0 \\ \\ ax^2+bx+c=0^{}^{} \\ a=-16,b=130,c=-201 \\ \\ t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t_1=\frac{-130+\sqrt[]{130^2-4\cdot(-16)\cdot(-201)}}{2\cdot(-16)} \\ t_1=\frac{-130+\sqrt[]{16900-12864}}{-32}=(-130+63.53)/(-32)=2.08\text{ seconds} \\ t_2=(-130-63.53)/(-32)=6.05\text{ seconds} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8wcbbdtl2igg551g4m3kv62qr9hmue4ail.png)
The firework hits this distance two times, the first time is when the firework is going upwards, and the time (rounding to the nearest tenth) is 2.1 seconds.