Given function is [tex]s(t)=-15t^2+52t+6[/tex]
where t is the time in seconds and s(t) is the height of the firework in t seconds.
Given that height of the firework is 45 feet. Now we have to find when it happens that is find value of t when s(t)=45
[tex]s(t)=-15t^2+52t+6[/tex]
[tex]45=-15t^2+52t+6[/tex]
[tex]15t^2-52t-6+45=0[/tex]
[tex]15t^2-52t+39=0[/tex]
Apply quadratic formula
[tex]t=\frac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\frac{-\left(-52\right)\pm\sqrt{\left(-52\right)^2-4\left(15\right)\left(39\right)}}{2\left(15\right)}[/tex]
[tex]t=\frac{52\pm\sqrt{2704-2340}}{30}[/tex]
[tex]t=\frac{52\pm\sqrt{364}}{30}[/tex]
[tex]t=\frac{52\pm 19.079}{30}[/tex]
[tex]t=\frac{52+19.079}{30}[/tex] or [tex]t=\frac{52-19.079}{30}[/tex]
[tex]t=\frac{71.079}{30}[/tex] or [tex]t=\frac{32.921}{30}[/tex]
[tex]t=2.3693[/tex] or [tex]t=1.09737[/tex]
Hence final answer is 2.3693 seconds and 1.09737 seconds.
Two answers means height of 45 feet will be attained two times.
