M1 question help!! DESPERATE!!!!!!!!!!!!!

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Hiya guys,

I'm kinda stuck on this M1 question: Kinematics

A particle is projected upwards with a speed of 14 ms-1. Find for how long it is above 2m 

I know its a really easy question but i seriously cannot get my head around it! lol =D

x

Modified once, last modified by Stacy on Mon 18th April, 2011 @ 19:15

Posted: 18-04-11 18:34 by Stacy

Is there any more information provided?

cuz I got:

u= 14m/s

s= 2m

and t=?

You'll need one more information to solve it. Like the acceleration (a) or final velocity (v).

Posted: 20-04-11 12:07 by Hanis94

welll... the acceleration is -9.8 cuz its being thrown upwards defying the gravity but thats about it .... :)

Posted: 20-04-11 19:42 by Stacy

anything to do with intergration?

Posted: 20-04-11 22:07 by Pui Pui

okay so I used the equation s = ut + 1/2at^2 and by the end of it I got a quadratic equation 

0 = 4.9t^2 - 14t + 2

In this case 't' would have two answers, and you would need to use x = (-b +-√ (b^2 – 4ac)) /2a

replacing x with t to find t. And that's where I'm stuck. You'd get two answers, and you wouldn't know which answer is the right one.....

Posted: 21-04-11 09:41 by Hanis94

SUVAT EQUATIONS WILL DO THE JOB :)

Posted: 22-04-11 14:56 by Georgina

u=14m/s v=0m/s a=-9.8m/s^2 find s

v^2 =u^2+2as

0=14^2+2x-9.8xs

0=196-19.6s

19.6s=196

s=10m (distance from start to highest point)

find v at s=2m   u=14m/s    a=-9.8m/s^2

use previous equation

v^2= 14^2 +2x -9.8 x 2

v^2= 196- 39.2

v^2 = 156.8

so the v at s=2m = 12.5

from 2m to 10m s=8m   u= 12.5m/s   v= 0m/s   a= -9.8m/s^2   find t

s= (u+v)/2 x t

8m = (12.5 + 0)/2 x t

8/6.25= t

t=1.28s

now the particle is moving backwards

so u= 0m/s  a= 9.8m/s^2 s= -8m

s=ut+1/2at^2

-8=0t +1/2 x 9.8 x t^2

-8= 1/2 x 9.8 x t^2

0= -4.9t^2+8

4.9t^2=8

t^2=12.9

t=3.59s

add together the two times:  1.28+3.59 = 4.87s

does this look right?

Posted: 22-04-11 15:56 by Pui Pui

Wow. Yeah I think so. :P

Posted: 22-04-11 16:05 by Hanis94

thanks took me ages to work it out.

TBH took me ages to understand the question and that it was in two parts

Posted: 22-04-11 16:15 by Pui Pui

Yeah, I didn't realize the question had a long process to it.  :P

Posted: 22-04-11 16:16 by Hanis94

=[ sorry guys book says its 2.56s, this thing is crazy! lol thanks a lot for helping out though :D

x

Posted: 25-04-11 07:16 by Stacy

That was one of my answers. I just didn't know which was the correct one since it was a quadratic. :) 

No problem :D

Posted: 25-04-11 08:10 by Hanis94

.:HanisMaggi:. wrote:

okay so I used the equation s = ut + 1/2at^2 and by the end of it I got a quadratic equation 

0 = 4.9t^2 - 14t + 2

In this case 't' would have two answers, and you would need to use x = (-b +-√ (b^2 – 4ac)) /2a

replacing x with t to find t. And that's where I'm stuck. You'd get two answers, and you wouldn't know which answer is the right one.....

Posted: 02-05-11 16:28 by Arron Beckett

Hanis,

take your equation and solve it - it is correct.  Use the quadratic formula and you get two answers, as you said, the smallest one (0.15) represents the time it takes to reach 2m on the way up. The other (2.71) represents the time it takes to go from the start, up to the max height and then back down to 2m.  If you take the two away from each other you get 2.56. the correct answer.  hope it makes sense.

Posted: 02-05-11 16:31 by Arron Beckett

i am a little too younng for this kind of question :P

Posted: 02-05-11 19:48 by Chloe

Oh yeah! Thanks Arron. :)

Posted: 02-05-11 20:35 by Hanis94