Mathematics: Afirework is shot into the air from an initial height of 6 feet. the height in feet of the firework above the ground is given by s(t)=-15t^2+52t+6, where t is time, in seconds, and t≥0, at what time will the firework be 45 feet off the ground?

The answer would be (7,-1) :)...

Afirework is shot into the air from an initial height

Alisewilcox

11.03.2022

Given function is s(t)=-15t^2+52t+6

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

s(t)=-15t^2+52t+6

45=-15t^2+52t+6

15t^2-52t-6+45=0

15t^2-52t+39=0

Apply quadratic formula

$t=\frac{-b \pm\sqrt{b^2-4ac}}{2a}$

$t=\frac{-\left(-52\right)\pm\sqrt{\left(-52\right)^2-4\left(15\right)\left(39\right)}}{2\left(15\right)}$

$t=\frac{52\pm\sqrt{2704-2340}}{30}$

$t=\frac{52\pm\sqrt{364}}{30}$

$t=\frac{52\pm 19.079}{30}$

$t=\frac{52+19.079}{30}$ or $t=\frac{52-19.079}{30}$

$t=\frac{71.079}{30}$ or $t=\frac{32.921}{30}$

t=2.3693 or t=1.09737

Hence final answer is 2.3693 seconds and 1.09737 seconds.

Two answers means height of 45 feet will be attained two times.

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