Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

An object is dropped from 39 feet below the tip of the pinnacle atop a ​715-ft tall building. The height of the object after seconds is given by the equation . h=16T^2+676.Find how many seconds pass before the object reaches the ground.

Sagot :

edisonlara1212

Answer:

6.5 seconds

Step-by-step explanation:

Ok so the equation you gave is -16T^2 + 676 in the comments which is what I'll be using for this problem. The problem is really just asking you to find the zeroes of the equation excluding any negative solutions since that doesn't really represent anything in this context since T is time.

So the first step is to set the equation equal to 0

[tex]0 = -16t^2 + 676[/tex].

Subtract 676 from both equations.

[tex]-676 = -16t^2[/tex]

Divide both sides by -16

[tex]42.25 = t^2[/tex]

Take the square root of both sides

[tex]\pm6.5 = t[/tex]

Ignore the negative and take only the positive solution, since in this context it doesn't make much sense. So after 6.5 seconds the height is 0, meaning it hits the ground after 6.5 seconds.

Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.