A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
A Cannon shell is fired straight up from the ground at an initial speed of 225 m/s. after how much time is the shell at a height of 620 m above the ground and moving down.
anonymous
 4 years ago
A Cannon shell is fired straight up from the ground at an initial speed of 225 m/s. after how much time is the shell at a height of 620 m above the ground and moving down.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I attempted to use the equation \[time=\frac{distance}{speed}\] but it didn't return an option that was a choice in the book, so i guess i need to do something else here.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0take into account that acceleration due to gravity is 9.81 and the question is asking when the ball is at a height of 620 m and ON ITS WAY DOWN. which means its a 2 part question. you have to find out how long it takes the ball to slow down to 0 speed from 225 at a negative acceleration of 9.81 then how much time it takes for the ball to get back to a height of 620 m with a positive acceleration of 9.81.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright let me try to work this out, and thanks for the help.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can use a kinematic equation here to find out these parts. first part is to find out where the ball stops and turns around. state your known values, which are acceleration= 9.81. Velocity initial = 225. Velocity Final= 0. so use the equation \[V ^{2}=V _{0}^{2}+2a* \Delta X\] and solve for \[\Delta X\] to find out how far the ball traveled up. then you can use another kinematic to find out how long that took. \[\Delta X = .5*(V _{0} + V) \Delta t\] by solving for \[\Delta t\] then you have to solve for the second part where its falling. this time your values are slightly different. we know our Velocity initial is 0. our delta x is our max height minus 620m and our acceleration is positive 9.81.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I suppose the answer is about 42s, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.043 sec is the an option so it sounds right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is 19.99s an option too? because thats what i got

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0They are 2.96, 17.3, 25.4, 33.6, 43.0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its 43 because i forgot to add my time from part 1 to part 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The shell is moving down, so we should add the time the shell take to reach the highest position

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But I don't see why there are so many answers...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0they are just giving me some to pick from

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, then surely is 43s

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0multiple choice. yeah im almost 100% sure it should be 43 s because it took about 23 s to reach max height then it took about 19.99 or 20 s to get back down to 620m

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Right on the target ,dude.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks a ton for the help!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no problem. you should be able to use the same method for your arrow question you posted

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yea i'm going to look at it again in the morning and see what i can't figure out. This might just be a case of me being brain dead right now.
Ask your own question
Sign UpFind 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.