Question

PLEASEEEEEEE HELPPPPPP ON ASSESSMENT PRACTICE ITS URGENT!!!HELP IS MUCH APPRECIATED!!!NOT GRADED!!!

Answer 1

The height of the firework is given by:

`y=-16x^2+80x+3`

where y represents the height and x the time.

We know that the velocity of the firework is the derivative of its position, then:

`v=(dy)/(dx)`

Then:

`\begin{gathered} v=(d)/(dx)(-16x^2+80x+3) \\ =-32x+80 \end{gathered}`

We also know that in its peak height the velocity of the firework will be zero, then we equate the velocity to zero and solve for x:

`\begin{gathered} -32x+80=0 \\ -32x=-80 \\ x=(-80)/(-32) \\ x=2.5 \end{gathered}`

Then, the firework reaches its peak height after 2.5 seconds. Once we know the times it takes the firework to reach this we plug this time into the equation for the height:

`\begin{gathered} y=-16(2.5)^2+80(2.5)+3 \\ =103 \end{gathered}`

Therefore, the maximum height of the firework is 103 feet.