A ball is thrown from a height of 160 feet with an initial downward velocity of 12f(t)/(s). The ball's height h (in feet) after t seconds is given by the following equation: h = 160 - 12t - 16t^2.
Added by Julia C.
Step 1
Step 1: We're given that the height h (in feet) of the ball after t seconds is: h = 160 - 12t - 16t² Show more…
Show all steps
Close
Your feedback will help us improve your experience
Qbs Educator and 91 other Calculus 1 / AB 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
If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet after t seconds is given by y = 40t - 16t^2. Find the velocity when t = 1.
Qbs E.
A ball is thrown straight up from the top of a 320-foot building with an initial velocity of 16 feet per second. The height of the ball over time is given by s = −16t^2 + 16t + 320 where s is measured in feet and t is measured in seconds. (a) Find the time when the ball reaches the ground (b) Determine the velocity of the ball when it reaches the ground.
Dominique Jan T.
A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance $s$ (in feet) of the ball from the ground after $t$ seconds is $s(t)=80 t-16 t^{2}$
Linear and Quadratic Functions
Inequalities Involving Quadratic Functions
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD