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Physics 8 Online
OpenStudy (anonymous):

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.

OpenStudy (anonymous):

I 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.

OpenStudy (anonymous):

take 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.

OpenStudy (anonymous):

alright let me try to work this out, and thanks for the help.

OpenStudy (anonymous):

you 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.

OpenStudy (anonymous):

I suppose the answer is about 42s, right?

OpenStudy (anonymous):

43 sec is the an option so it sounds right.

OpenStudy (anonymous):

is 19.99s an option too? because thats what i got

OpenStudy (anonymous):

They are 2.96, 17.3, 25.4, 33.6, 43.0

OpenStudy (anonymous):

its 43 because i forgot to add my time from part 1 to part 2

OpenStudy (anonymous):

The shell is moving down, so we should add the time the shell take to reach the highest position

OpenStudy (anonymous):

But I don't see why there are so many answers...

OpenStudy (anonymous):

they are just giving me some to pick from

OpenStudy (anonymous):

Ah, then surely is 43s

OpenStudy (anonymous):

multiple 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

OpenStudy (anonymous):

Right on the target ,dude.

OpenStudy (anonymous):

Thanks a ton for the help!

OpenStudy (anonymous):

no problem. you should be able to use the same method for your arrow question you posted

OpenStudy (anonymous):

Yea 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.

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