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# Maths Mechanics 2 Help! watch

1. Hi could any1 of u help me in solving this problem..thank u
Question
A particle slides along a horizontal table and over the edge. Given that the table is 1m high and the speed of the particle on the table is 14ms^-1 calculate:
(a) the vertical velocity with which the particle hits the floor(assumed horizontal)
(b) the speed and direction of motion of the particle when it hits the floor.
2. What have you got so far?
3. (Original post by Yiikes010)
Hi could any1 of u help me in solving this problem..thank u
Question
A particle slides along a horizontal table and over the edge. Given that the table is 1m high and the speed of the particle on the table is 14ms^-1 calculate:
(a) the vertical velocity with which the particle hits the floor(assumed horizontal)
(b) the speed and direction of motion of the particle when it hits the floor.
For a. the vertical speed at the start is 0. so U=0. a=9.8 etc...

equations of motion.

the horizontal component will not change. so your answer so (a) and 14.. use pythagoras (can't spell) to work the resultant speed. and trig to get the angle.
4. For a, use

For b, use your answer to a, and the horizontal velocity given, and do some pythagoras and trig.
5. k.. dis is what i did
Vertical comp.
v^2 = u^2 +2as..and found out what v is..bt it is not the right answer!(btw u=0)
6. (Original post by Yiikes010)
k.. dis is what i did
Vertical comp.
v^2 = u^2 +2as..and found out what v is..bt it is not the right answer!(btw u=0)
didn't you get...

v^2 = 19.6
7. (Original post by Pork and Beans)
For a, use

For b, use your answer to a, and the horizontal velocity given, and do some pythagoras and trig.
I did tat eq for a but it doesnt give the answer..its menat 2 b 7.8ms^-1..i got 4,3ms^-1
8. (Original post by Yiikes010)
k.. dis is what i did
Vertical comp.
v^2 = u^2 +2as..and found out what v is..bt it is not the right answer!(btw u=0)

Everything you've said so far seems correct.
9. (Original post by kam_007)
didn't you get...

v^2 = 19.6
yep i did
10. You got the answer to a right (well, close. I got 4.43).
11. (Original post by ghostwalker)

Everything you've said so far seems correct.
k.. i took u=0, a=9.8(+ve downwards) and s=1 and put them in the eq and gt 4.4 ms^-1...bt apparently its wrong for part a
12. (Original post by Pork and Beans)
You got the answer to a right (well, close. I got 4.43).
oh yeh i got 4.43..missed d 4 i guess lol
13. For the question, as you have posted it, 4.43 is the correct vertical velocity with which is hits the floor.

And please stop using text speak, it's banned on this site, and your post are barely intelligable.
14. Isn't this Mechanics 1 stuff..?

Or maybe I've completely forgotten whats on this years course, very possible.
15. (Original post by ghostwalker)
For the question, as you have posted it, 4.43 is the correct vertical velocity with which is hits the floor.

And please stop using text speak, it's banned on this site, and your post are barely intelligable.
thank you for letting me know..i am new to tsr!
16. (Original post by laura_beth)
Isn't this Mechanics 1 stuff..?

Or maybe I've completely forgotten whats on this years course, very possible.
no its not according to edexcel maths..its the first chapter
17. It looks like the answer given in the book is incorrect for part a, so it's also unlikely to be correct for part b (but you never know).

Are you happy with being able to do part b?
18. so why does the back of my book says the answer is 7.8ms^-1?did i do something wrong?
19. and can anyone help me with part b..i dont know how to find the speed!!
20. (Original post by Yiikes010)
no its not according to edexcel maths..its the first chapter
ahh I do AQA possibly the reason

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