We are standing on the top of a 960 feet tall building and launch a small object upward. The object's vertical position, measured in feet, after t seconds is h(t) = -16t^2 + 64t + 960. What is the highest point that the object reaches?
Added by Cathy M.
Close
Step 1
Step 1: The equation for the object's vertical position is h(t) = 16t^2 + 64t + 960. Show more…
Show all steps
Your feedback will help us improve your experience
Supreeta N and 58 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Vishal P.
We are standing on the top of a 304 feet tall building and launch a small object upward. The object's vertical position, measured in feet, after t seconds is h(t) = -16t^2 + 288t + 304. What is the highest point in feet that the object reaches?
Kathleen C.
Use this information: At a time $t$ seconds after an object is tossed vertically upward, it reaches a height s in feet given by the equation $s=80 t-16 t^{2}$. What is the maximum height reached by the object?
Functions and Function Notation
Quadratic Functions
Recommended Textbooks
Elementary and Intermediate Algebra
Algebra and Trigonometry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD