Question

One of the fireworks is launched from the top of the building with an initial upward velocity of 150ft/sec. When will the firework land? The equation of the fireworks path is h(t) =-16t^2+150t +50

Answer 1

Answer:

9.70 seconds

Explanation:

The function that represents the path of the fireworks is:

$h\mleft(t\mright)=-16t^2+150t+50$

The object will land when the height, h(t)=0.

$-16t^2+150t+50=0$

We solve the equation for t using the quadratic formula.

$\begin{gathered} a=-16,b=150,c=50 \\ \textcolor{red}{t=\dfrac{-b\pm\sqrt[]{b^2-4ac}}{2a}} \\ t=\dfrac{-150\pm\sqrt[]{150^2-4(-16)(50)}}{2\times-16}=\dfrac{-150\pm\sqrt[]{22500+3200}}{-32} \\ =\dfrac{-150\pm\sqrt[]{25700}}{-32} \\ t=\dfrac{-150+\sqrt[]{25700}}{-32}\text{ or }t=\dfrac{-150-\sqrt[]{25700}}{-32} \\ t\neq-0.322\; \text{or }t=9.697 \end{gathered}$

The fireworks will land after 9.70 seconds.