A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
16t^2+vt+s
A ball is thrown straight down from the top of a 220foot building with a velocity of 22feet per second. What is its velocity after falling 108 ft?
 2 years ago
16t^2+vt+s A ball is thrown straight down from the top of a 220foot building with a velocity of 22feet per second. What is its velocity after falling 108 ft?

This Question is Closed

cyter
 2 years ago
Best ResponseYou've already chosen the best response.0\[\huge v_{f}^2 = v_{i}^2+2ad\] d=108ft a=9.8m/sec^2 vi=22ft/sec

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1s(t)=16t^222t+220 This function measures how far above the ground the ball is, so when it falls 108 ft, it is 112 ft above the ground s(t)=16t^222t+220=112 s(t)=16t^222t+108=0 Using quadratic equation to solve for t: \[t=\frac{22 \pm \sqrt{4844(16)(108)}}{2(16)}=\frac{22 \pm 86}{16}\] We can discard the 22+86 since it will yield a negative time. So t=(2286)/32=2 v(t)=s'(t)=32t22 v(2)=32(2)22=6422= 86ft/s
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.