Question

Please help!!

A firework 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?

Answer 1

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=(-b \pm√(b^2-4ac))/(2a)`

`t=(-\left(-52\right)\pm√(\left(-52\right)^2-4\left(15\right)\left(39\right)))/(2\left(15\right))`

`t=(52\pm√(2704-2340))/(30)`

`t=(52\pm√(364))/(30)`

`t=(52\pm 19.079)/(30)`

`t=(52+19.079)/(30)` or
`t=(52-19.079)/(30)`

`t=(71.079)/(30)` or
`t=(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.