Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second.
y=-16x^2+161x+108
y=−16x
2
+161x+108

Sagot :

abidemiokin

Answer:

5.03secs

Step-by-step explanation:

Given the height reached by the rocket expressed as;

y = -16x^2 + 161x + 108

The rocket reaches it max height when its velocity is zero

Velocity is expressed as;

V(x) = dy/dx  = 0

dy/dx = -32x + 161

0 = -32x + 161

32x = 161

x = 161/32

x = 5.03

Hence it will take 5.03secs for the rocket to reach its maximum height

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.