Question

1. fireworks are discharged vertically upward. its path is given by the equation: $h(t)=-16t^{2}+160t + 4$, where $h(t)$ is the height of the fireworks in feet and $t$ is the time in seconds.

a) when will the firework reach its maximum height? what is that height?
b) what is the initial height of the firework?
c) when is the firework 50 feet above the ground?
d) if the firework were to keep going, when will it touch the ground?

Explanation:

Step1: Identify the coefficients

The height - time function is $h(t)=-16t^{2}+160t + 4$, where $a=-16$, $b = 160$, $c = 4$.

Step2: Find the time of maximum height

For a quadratic function $y = ax^{2}+bx + c$, the time $t$ at which the maximum (since $a<0$) occurs is given by $t=-\frac{b}{2a}$.
$t=-\frac{160}{2\times(-16)}=\frac{160}{32}=5$ seconds.

Step3: Find the maximum height

Substitute $t = 5$ into the height - function $h(t)$:
$h(5)=-16\times5^{2}+160\times5 + 4=-16\times25+800 + 4=-400+800 + 4=404$ feet.

Step4: Find the initial height

The initial height is when $t = 0$. Substitute $t = 0$ into $h(t)$:
$h(0)=-16\times0^{2}+160\times0 + 4=4$ feet.

Step5: Find when $h(t)=50$

Set $h(t)=50$, so $-16t^{2}+160t + 4 = 50$.
Rearrange to get $-16t^{2}+160t-46 = 0$. Divide through by $-2$: $8t^{2}-80t + 23 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a = 8$, $b=-80$, $c = 23$.
$t=\frac{80\pm\sqrt{(-80)^{2}-4\times8\times23}}{2\times8}=\frac{80\pm\sqrt{6400-736}}{16}=\frac{80\pm\sqrt{5664}}{16}=\frac{80\pm75.26}{16}$.
$t_1=\frac{80 + 75.26}{16}\approx9.70$ seconds and $t_2=\frac{80 - 75.26}{16}\approx0.30$ seconds.

Step6: Find when $h(t)=0$

Set $h(t)=0$, so $-16t^{2}+160t + 4 = 0$. Divide through by $-4$: $4t^{2}-40t - 1=0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a = 4$, $b=-40$, $c=-1$.
$t=\frac{40\pm\sqrt{(-40)^{2}-4\times4\times(-1)}}{2\times4}=\frac{40\pm\sqrt{1600 + 16}}{8}=\frac{40\pm\sqrt{1616}}{8}=\frac{40\pm40.2}{8}$.
$t=\frac{40+40.2}{8}\approx10.03$ seconds (we take the positive root since time cannot be negative).

Answer:

a) The firework reaches its maximum height at $t = 5$ seconds and the maximum height is $404$ feet.
b) The initial height of the firework is $4$ feet.
c) The firework is 50 feet above the ground at approximately $t\approx0.30$ seconds and $t\approx9.70$ seconds.
d) The firework touches the ground at approximately $t\approx10.03$ seconds.