looks like you need a little bit of trig and the equations for accelerated motion

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quote:

Originally posted by Zild looks like you need a little bit of trig and the equations for accelerated motion

Time to get super-dooper creative and go where no one else has been before!

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its a bit trickier then that. You don't have the angle for the second stone all you have is the initial velocity and the time


you need another equation to equate the two.

quote:

Originally posted by Nrg2Nfinit its a bit trickier then that. You don't have the angle for the second stone all you have is the initial velocity and the time you need another equation to equate the two.

if you have the time you have the height then you have the angle

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the stone is thrown off the top.. let me draw a diagramn.. The height of the building wont really help unless you know the angle at which it is thrown.






its going up (past the height) then down again.

quote:

Originally posted by Nrg2Nfinit i couldnt figure this one out on my exam .. it was tough lets see if anyone can do it here. 2 stones are thrown from the top of a building at 35 m/s. The first one lands after 5 seconds and travels a distance of 150 meters. The second one lands 2 seconds after. Find. Both initial angles The height of the building The distance for the second stone. Good luck I can get all the information for the first one.. but i'm lost with the second one completely

I fail at math

240.35 meters high

quote:

Originally posted by Zild looks like you need a little bit of trig and the equations for accelerated motion

yeah, even i could figure that one out. well, if i was still in highschool and remembered which equations did what.

quote:

Originally posted by Domesticated Time to get super-dooper creative and go where no one else has been before!

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here i posted what i solved for



x=xo + vot (no acceleration in x component)

1)y=yo+vot +1/2 at^2
2)v=vo+at (you can use this for y component)
3)v^2=vo^2+ 2ay (you can use this for y component as well)


so for the first part i got the angle by using tvocos@=x

since i have x t and v i can solve for @ which gives 31 degrees.

From there i found my y componet by first solving for distance to zero velocity which gave 15.5 meters (v2=v02+2ay) where v=0 a -9.81.

I also solved for how much time that took using v=vo+at which gave me 1.78 seconds. I can subtract this from 5 to give me the remaining time that it will take to hit the ground from maximum height (zero velocity) 3.21.

I can use y=yo + 1/2at2 to solve for the distance to hit the ground from zero velocity. I subtract this number (50.7) from 15.5 and i get the height of the building to be 35.2.


This is all for the first stone (black) now all i know for the second stone (red) is that it lands 2 seconds after the first stone and it has the same velocity (35m/s).


Good luck

quote:

Originally posted by pkcRAISTLIN yeah, even i could figure that one out. well, if i was still in highschool and remembered which equations did what.

Yeah, this.

You are fucking dumb if you're doing this stuff at school with a text book and can't work it out. Physics is one of the easiest kinds of maths, at least at this level.

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quote:

Originally posted by Sunsnail 240.35 meters high

2 seconds later.. not 500 seconds later. and its the same velocity as the first one.