A ball is thrown vertically upward with an initial velocity of 64 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s equals 64t - 16t^2. The ball is more than 48 feet above the ground for the time t when 0 < t < 4.
Added by Esperanza J.
Step 1
We know that the ball is more than 48 feet above the ground, so we can set up an inequality: s > 48 Show more…
Show all steps
Close
Your feedback will help us improve your experience
Daniel Carr and 82 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
A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance $s$ (in feet) of the ball from the ground after $t$ seconds is $s(t)=96 t-16 t^{2}$. (a) At what time $t$ will the ball strike the ground? (b) For what times $t$ is the ball more than 128 feet above the ground?
Linear and Quadratic Functions
Inequalities Involving Quadratic Functions
A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance $s$ (in feet) of the ball from the ground after $t$ seconds is $s(t)=96 t-16 t^{2}$ (a) At what time $t$ will the ball strike the ground? (b) For what time $t$ is the ball more than 128 feet above the ground?
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 $s(t)=80 t-16 t^{2}$. (a) At what time $t$ will the ball strike the ground? (b) For what time $t$ is the ball more than 96 feet above the ground?
Recommended Textbooks
Elementary and Intermediate Algebra
Algebra and Trigonometry
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD