Recent content by Jai

Thanks, I applied them separately and managed to get what I need. While you are here do you know some sort of formula that could be used to take into account air resistance? My teacher said that we need to mention the formula but not calculate it because it is too complicated to get the data for.

Yes, however I mean that for my original question, so now the approach may be different?

Yes, since a bullet is being shot out of a gun.

Oh wow. Just realized something... 15.6 m/s is the velocity and 0 is the initial velocity... Sorry about that...

No I don't.

None at all unfortunately.

Well I guess since we are neglecting any vertical forces a is going to equal 0 since there is no air resistance and u will equal the horizontal component which is 15.068.

I didn't realize that thanks.

So the time would be equal?

"a" would be 9.8m/s^2 and "u" is 15.6 m/s

Maybe not since it would be decellerating while traveling up and accelerating while coming down?

Yes I could use s=ut+1-2at^2 to calculate that.

Yes I understand what you mean. Would it depend on the velocity it was thrown at since it will constantly accelerate as it comes back down, meaning that if it has more time to do so it will end up at a greater speed?

Yes, it is launched from ground level. Won't the differences in time depend on the angle?

Possibly. I did work something out but I am not sure if it is correct. the initial vertical velocity is 4.038 m/s^2. I then subtracted the acceleration due to gravity giving me -5.77m/s^2. Sorry if this is all wrong I've just really confused myself here...