Problem sets for MACE-11010 Engineering Mechanics

I'm Karl Conlon on Aerospace Engineering

give me the points haha

A + B = (1-1,1+1,0+0) = (0,2,0)

A - B = (1+1,1-1,0-0) = (2,0,0)

2A + 3B = (2-3,2+3,0+0) = (-1,5,0)

A.B = (1x-1,1x1,0x0) = (-1,1,0)

i think lol

hi i am Rameez.M...doin mech eng

v^2=u^2+2as a=9.81/6=1.635  v^2=2^2+(2x1.635x5)=18.35   v=4.51m/s

Hi im Alex Proctor n im also doin aerospace engineering

how do you keep track of your points? are they automatically given to you or do you have to keep note yourself? if you track them yourself how do you make sure people are bein honest

Hi I'm Sam Ratcliffe and I'm also doing Aerospace

The cross product one:

AxB = (0,0,2)

hey i'm Henry Stewart. i'm doing mech eng . heres my answers for the questions

1) A+B = 0,2,0

2) A-B = 2,0,0

3) 2A + 3B = -1,5,0

4) A.B = 0,0,0

5) AxB = -1,1,0 

Q2) i got the same as kean for this one and can't see where he went wrong

V^2 = U^2 + 2AS

V = SQR(2^2 + 2x(9.81/6)x5)

V = SQR(20.35)

V = 4.51m/s 

Does that make me the first to post all the answers (even if they are wrong) 

Question 1

2-  0.2.0

3-  2.0.0

4-  -1.5.0

5-  0

6-  0-0.0-0.1-(-1) = 0.0.2

Question 2:

5=1/2*1/6*9.81t^2+2t

t=1.54  -->  V=9.81/6*1.54+2   ---> V=4.52

Question 1 - Vectors

A+B=(0,2,0)

A-B=(2,0,0)

2A+3B=(-1,5,0)

A.B=0

AxB=(0,0,2) (Assuming Cross Product)

Question 2 - Kinematics

Initial Velocity = 2 m/s, Acceleration = 9.81/6 m/s^2, Distance = 5m, Final Velocity = V

V^2=(Initial Velocity)^2 + 2(Acceleration x Distance)

V^2= 4 + 2((9.81/6) x 5)

V^2= 20.35

V = 4.51 m/s

10 Points Henry?? 

Question 1

2-  0.2.0

3-  2.0.0

4-  -1.5.0

5-  0

6-  0-0.0-0.1-(-1) = 0.0.2

Question 2:

5=1/2*1/6*9.81t^2+2t

t=1.54  -->  V=9.81/6*1.54+2   ---> V=4.52

A+B=(1,1,0)+(-1,1,0)=(0,2,0)

A-B=(1,1,0)-(-1,1,0)=(2,0,0)

2A+3B=2(1,1,0)+3(-1,1,0)=(-1,5,0)

A.B= (1,1,o).(-1,1,0)=0

Cross product of AB=(0,0,2)

And i don't support your system of evaluating students as; this is not the proper way. I hope University does not allow you to play havoc with its students.

Q1:

 A+B = (1 - 1, 1 + 1, 0 + 0) = (0,2,0)

 A-B = (1 + 1, 1 - 1, 0 - 0) = (2,0,0)

2A + 3B = (2 - 3, 2 + 3, 0 + 0) = (-1,5,0)

A.B = 1.-1 + 1.1 + 0.0 = 0

AxB = (1.0 - 0.1) - (1.0 - 0.-1) + (1.1 - 1.-1) = (0,0,2)

Q2:

v^2 = u^2 + 2as 

u = 2m/s

a = 9.81/6m/s/s

s = 5m

Therefore:

v^2 = 4 +  16.35 = 20.35

v = 4.51m/s

But everyone has already posted this so...? 

I'm doing aersopace eng

I know lots of people have already posted their answers but I may aswell also post them. 

Q1

A + B

(1) + (-1) = 0
(1) + (1) = 2
(0) + (0) = 0

(0,2,0)

A-B

(1) - (-1) = 2
(1) - (1) = 0
(0) - (0) = 0

(2,0,0)

2A+3B

2(1) + 3(-1) = -1
2(1) + 3(1) = 5
2(0) + 3(0) = 0

(-1,5,0)

A.B

(1)*(-1) = -1
(1)*(1) = 1
(0)*(0) = 0

1-1=0

A*B

(1)*(-1) = -1
(1)*(1) = 1
(0)*(0) = 0

(-1,1,0)

Q2

U= 2 ms^-1
S= 5 m
a= 9.81/6 ms^-2

Finding V.

Use

v^2=u^2+2as
v^2= 4+2*1.635*5
v= 4.51 ms^-1

hi there... I know other ppl have posted their answers, but that shouldn't be a reason for me not to do it, rite? so, below are my working and answers.

Q1:

A = (1,1,0)
B = (-1,1,0)

A + B = (0,2,0)

A - B = (2,0,0)

2A + 3B = (2,2,0) + (-3,3,0) = (-1,5,0)

A . B = (1 x -1) + (1 x 1) + (0 x 0) = -1 + 1 + 0 = 0

A x B = [(1x0)-(1x0)]i - [(1x0)-(-1x0)]j + [(1x1)-(-1x1)]k
        = (0,0,2)

Q2:

u = 2 m/s ; s = 5m ; a = (9.81/6) m/s^2 ; v = ?

v^2 = u^2 + 2as
v^2 = (2)^2 + 2(9.81/6)(5)
v^2 = 20.35

v = 4.511 m/s 

Hi my name is Farhan bin Zainol, currently in Mechanical Engineering (BEng) course.

Here are my answers to the questions.

Q1:

Since A = (1, 1, 0) and B = (-1, 1, 0)

--> A + B = (0, 2, 0)

--> A - B = (2, 0, 0)

--> 2A + 3B = (-1, 5, 0)

 --> A.B = (1*-1) + (1*1) + (0*0)

            = 0

--> A x B is a cross-product. Hence,

A x B = [(1*0) - (1*0)] i - [(1*0) - (1*0)] j + [(1*1) - (-1*1)] k

           = 2 k.

Therefore, A x B = (0, 0, 2)

Q2:

Initial velocity = 2 m/s

Height = 5 m

Acceleration = (9.81/6) m/s²

By using v² = u² + 2as,

v² = (2)² + 2 (9.81/6) (5)

    = 20.35

v   = 4.51 m/s  (3 significant figures)

I get the same as everyone else for question 2 and dont understand how it could b wrong

v² = u² + 2as,

v² = 2² + 2 (9.81/6x5)

    = 20.35

v   = 4.51 m/s

Questions from Engineering Mechanics, tenth edition - by R.C Hibbeler.

1. Solve the following equation for x, y, and z:

x – y + z = –1   –x + y + z = –1   x + 2y – 2z = 5

x - y + z = -1 -- ( 1 )
-x + y + z = -1 -- ( 2 )
x + 2y - 2z = 5 -- ( 3 )

Eq. (1) + Eq. (2):
2z = -2
hence, z = -1

Eq. (2) + Eq. (3):
3y - z = 4
3y - -1 = 4
3y = 3
y =1

From Eq. (1):
x - y + z = -1
x -1 + -1 = -1
x = 1

2. Solve the following equation for the two roots of x:   x2 — 16 = 0

x² = 16
x = ±√16
x = +4, -4

3.
Solve the following equation for the two roots of x:   — x2 + 5x = — 6 

 x² - 5x -6 =0
 (x-6)(x+1)=0
x = 6, x = -1

4.
Determine the angle :

 θ + 40° = 90°
 θ = 50°

5.
Determine the angle :

 θ + 40° = 90°
 θ = 50°

6.

Determine the length of side AB if right angle ABC is similar to right angle A'B'C': 

B'C' =  √(15² + 36²)
B'C' = 39

AB/A'B' = BC/B'C'

AB = A'B' x BC / B'C'
     =15 x 13/39
     = 5

7.
Using the basic trigonomic functions, determine the length of side AB of the right triangle. 

AB/AC = sin 30°
AB = 10x0.5
     = 5

8.
Using the basic trigonomic functions, determine the length of side AB of the right triangle. 

AB/AC = sin 45°
AB = 10 x (√2) / 2
    = 5√2
    = 7.071

9.
Using the basic trigonomic functions, determine the length of side AB of the right triangle. 

AB/BC = tan 45°
AB =  10 x 1
    = 10

10.
Determine the angles and
and the length of side AB of the triangle. Note that there are two
possible answers to this question and we have provided only one of them
as an answer.

6 / sin 40° = 7 / sin Φ
sin Φ = (7 / 6) sin 40°
        = 0.7499
Φ = 48.6°

40° + Φ +  θ = 180°
θ = 180° - 40° - 48.6°
  = 91.4°

AB / sin 91.4° = 6 / sin 40°
AB = 6 sin 91.4° / sin 40°
    = 9.33

1. Static's deals with bodies that have constant velocity.
- False, statics means not moving

2. Equilibrium means at rest or moving with constant velocity.
- True

3. A particle has a mass but its size can be neglected.
- True

4. The particles that compose a rigid body remain at a fixed distance from one another both before and after a load is applied.
- True

5. A concentrated force is assumed to act at a point on a body
- True

6. A particle subjected to a system of forces will remain at rest provided the force system is balanced.
- True

7. The weight of a body remains constant regardless of its location.
- False, it depends on the gravity force which depends on location.

8. The SI system considers force to be a derived unit.
- True

9. A newton is equal to the force required to accelerate 1 kg at 9.81 m/s2.
- False, it is the force to accelerate a mass of 1 kg at 1 m/s²

10. The prefix giga ( G ) indicates 109.
- True

11. The prefix mega ( M ) indicates 10-6.
- False, should be 10^6

12. The unit ms is a milli-second.
- True

13. A milli-meter squared is written as mm2.
- True

14. It is satisfactory to write k µ s as an appropriate composite unit.
- True

15. The set of composite units s/kg are allowed.
- True

16. The accuracy of a number is indicated by the number of significant figures it contains.
- True

17. 2.16 rounded off to two significant figures is 2.2.
- true

18. 2.25 rounded off to two significant figures is 2.3.
- True

19. 2.05 rounded off to two significant figures is 2.0.
- False, should be 2.1

20. 2.35 rounded off to two significant figures is 2.4.
- True

Hi! I'm Tee Soon Teck from Mechanical Engineering class (BEng).

Below are my solutions and answers:

Vector

 A + B = (i,j) + (-i,j) = 2j or (0,2,0)

A - B = (i,j) - (-i,j) = 2i or (2,0,0)

2A + 3B = 2(i,j) + 3(-i,j) = -i,5j or (-1,5,0)

A . B = (i.j) . (-i,j) = -1+1 = 0

A x B = [(1 x 0) - (1 x 0)]i - [(1 x 0) - (1 x 0)]j + [(1 x 1) - (1 x 1)]k = 2k or (0,0,2)

Kinematics

u = 2 m/s

v = ?

a = (9.81/6) m/s^2

s = 5 m

v^2 = u^2 + 2as

v^2 = 2^2 + 2 (9.81/6) (5)

v^2 = 20.35

v = 4.511 m/s

Okay, here's the notoriously difficult-to-solve but apparently easy Question number two. The strange thing is, I think all the above posts have the correct answer.

The descent engine stops at h = 5m, which means that it doesn't have any force acting on it, except for the lunar gravity. At that point of time, t, its velocity is u = 2m/s. Just before the landing gear impacts, (presuming that the landing gear is attached to the space vehicle) the final velocity will obviously be given by the formula v^2 = u^2 + 2ah.

 And hence, v^2 = 20.35 m^2/s^2 Hence, v = 4.51 m/s

But after it impacts the groud, its velocity will obviously be zero. Isn't that correct?

Hey, I'm Dave and I'm studying Aerospace Engineering. I am HUGELY sceptical of this assement system and honestly think it will never work! I've posted on another page (http://www.imechanica.org/node/1931#new) explaining all this - please go on and tell me if you agree (that's to all students as well as Mr Tan). I'm really not looking forward to a year of me being marked and assesed in this way!
Here are my answers, I did do them myself but I could have just as well copied from someone else - this system makes NO SENSE!

Q1 

A + B = (1 + -1), (1 + 1), (0 + 0) = (0,2,0)

A - B = (1 - -1), (1 - 1), (0 - 0) = (2,0,0)

2A + 3B = 2(1,1,0) + 3(-1,1,0) = (2,2,0) + (-3,3,0) = (-1,5,0)

A.B = (1 * -1) + (1 * 1) + (0 * 0) = -1 + 1 + 0 = 0

AxB = i[(1 * 0) - (0 * 1)] + j[(1 * 0) - (-1 * 0)] + k[(1 * 1) - (1 * -1)] = 0i + 0j + 2k = (0,0,2)

Q2

v^2 = u^2 + 2as
s = 5, a = 9.81/6 = 1.635, u = 2

v^2 = (2^2) + (2 * 1.635 * 5)
v^2 = 4 + 16.35
v^2 = 20.35
v = 4.51 m/s (3 s.f.)

Does anyone else find it take ages to type this stuff out, or do I just need more practice?

My name is Yu Wu, a aerospace engineering student

A+B=(1,1,0)+(-1,1,0)=(0,2,0)
A-B=(1,1,0)-(-1,1,0)=(2,0,0)
2A+3B=2(1,1,0)+3(-1,1,0)=(-1,5,0)
A.B=(1,1,0).(-1,1,0)=1*(-1)+1*1+0=0
AxB=| i j k |
| 1 1 0 |
|-1 1 0 |

=0-0+(1*1+1*1)k
Written as (0,0,2)

With the formula: 2as=v^2-u^2
so v=(2as+u^2)^(1/2)
=(2*9.81*5/6+2^2)^(1/2)
Approximiately=4.51m/s

Hi i'm chirag, i'm definitly late with these answers but oh well. 

Q1:

A + B = (0,2,0)

A - B = (2,0,0)

2A + 3B = (2,2,0) + (-3,3,0) = (-1,5,0)

A . B = (1 x -1) + (1 x 1) + (0 x 0) = -1 + 1 + 0 = 0

A x B = [(1x0)-(1x0)]i - [(1x0)-(-1x0)]j + [(1x1)-(-1x1)]k
        = (0,0,2)

Q2:

u = 2 m/s 

 s = 5m 

 a = (9.81/6) m/s^2 

 v = ?

v^2 = u^2 + 2as
v^2 = (2)^2 + 2(9.81/6)(5)
v^2 = 20.35

v = 4.511 m/s 

 Q1

A+B : (0,2,0)

A-B : (2,0,0)

2A+3B : (-1,5,0)

A.B : 0

AxB : (0,0,2)

 Q2

v=5m/s (1sf) 

A+B= (0,2,0)

A-B= (2,0,0)

2A= (2,2,0)  3B= (-3,3,0)       2A+3B= (-1,5,0)

A.B= (1x-1)+(1x1)+(0x0)=0

AxB= ((1x0)-(0x1)), ((0x-1)-(1x0)), ((1x1)-(1x-1))

     = 0,0,2

 
Q2

s=5, a=9.81/6, u=2, v=?

v^2=u^2+(2as)

v=4.51 m/s at the exact moment of impact after which the velocity will decrease to ZERO as the energy is disappated. 

I know I'm certainly not first to answer but it's been my first chance to get to a computer all day 

Q1

A+B = (0,2,0)

A-B = (2,0,0)

2A+2B = (2,2,0) + (-3,3,0) = (-1,5,0)

A.B = (1x-1) + (1x1) + (0x0)  = 0

AxB = ((-1x0)-(0x1)) , ((0x-1)-(1x0)) , ((1x1) – (1x-1)) = (0,0,2)

Q2

V^2 = u^2+2as            s=5 , a=9.81/6 = 1.635 ,  u=2

v^2 = (2^2)+(2x1.635x5) = 20.35

Therefore: v = 4.51 ms

Q1

A + B

(1) + (-1) = 0
(1) + (1) = 2
(0) + (0) = 0

A-B

(1) - (-1) = 2
(1) - (1) = 0
(0) - (0) = 0

2A+3B

2(1) + 3(-1) = -1
2(1) + 3(1) = 5
2(0) + 3(0) = 0

A.B

(1)x(-1) = -1
(1)x(1) = 1
(0)x(0) = 0

so -1+1+0=0

AxB

(1)x(-1) = -1
(1)x(1) = 1
(0)x(0) = 0

Q2

U= 2 m/s
S= 5 m
a= 9.81/6 m/s2

v2=u2+2as
v2= 4+2x1.64x5
v= 4.51 m/s 

here you go! 

Q1
A + B = (0,2,0)
A - B = (2,0,0)
2A + 3B = (2,2,0) + (-3,3,0) = (-1,5,0)
A . B = (1 x -1) + (1 x 1) + (0 x 0) = -1 + 1 + 0 = 0
A x B = (1x0)-(1x0) - (1x0)-(-1x0) + (1x1)-(-1x1) = (0,0,2)

Q2:
u = 2 m/s    s = 5m     a = (9.81/6) m/s^2    v = ?
v^2 = u^2 + 2as
v^2 = 2^2 + 2(9.81/6)5 
v = 4.511 m/s

Q1.

(A+B)= (1-1),(1+1),(0+0)= (0,2,0)

(A-B)= (1+1),(1-1),(0-0)= (2,0,0)

2A+3B= 2(1,1,0) + 3(-1,1,0)=(2,2,0)+(-3,3,0)= (-1,5,0)

A.B = (1x-1)+(1x1)+(0x0)=(-1+1+0)= 0

AxB = [(1x0)-(0x1)],[(1x0)-(-1x0)],[(1x1)-(1x-1)] = (0,0,2)
^This is the cross product. Some people seem to be confusing it with the scalar/dot product.

Q2.

v2 = u2 + 2as
u=2, a=9.81/6=1.635, s=5

v2 = 2x2 + 2x1.635x5
v2 = 20.35
v2 = 4.51 ms-1 (to 3 s.f.)

[edit] In response to Q3:

Newtons 2nd law states that, observed from an inertial reference frame, the net force on a particle is proportional to the time rate of change of its linear momentum. Apply this to a body and the simple equation is F=ma; F is the force needed to be applied to a body of mass m in order to produce an accceleration a.

Hi, I am Aman Mehra, Mechanical Engineering.

Here are my answers:

Q1:

A+B =  (1-1,1+1,0+0) = (0,2,0)

A-B =    {1-(-1),1-1,0-0} = (2,0,0)

2A + 3B = (2-3,2+3,0+0) = (-1,5,0)

A.B = (1*-1) + (1*1) + (0+0) = 0

A x B = {0-0}, {0-0}, {1-(-1)} = (0,0,2)

Q2:

s = 5m   u = 2m/s    a = 9.8/6    v = ?

Formula used: v² = u² + 2as

v² = 4 + 5(9.8/3)

v² = 61/3

v = 4.51 (3s.f.)

Regarding the system debate, I highly doubt this would do the students any good. I would prefer to go back to the traditional system.

Hey all, heres a little email i sent mr. Tan about what i think of the system:

 
Dear Prof. Tan,

Firstly, I would like to introduce myself, my name is Rafael, shortly
known as Raf. Secondly, I would like to comment on this method of
learning. At the same time of which I found it interestingly innovative, I
realized its great unfairness. Now that so many people have answered, one
could argue that there is no reason for me to answer, as it could seem
that I am simply copying the previous answer. Similarly, others could
always wait for the first answer to be posted, and gain points for copying
the right answer, and unfortunatelly, not study/learn. I believe that if
each one of us were to send u a personal message with our answer, it would
be fairer. Now, as everyone can see eachothers working, answering to the
blog is like doing a test with the answers; you can work them out, but you
dont have to, you can see them. I dont want to be negative, but I believe
that this system is fairly naive. Whereas, if we were all to mail you the
answer, and you reply with how many points we gain...it would be fair for
everyone, specially people who have activities, and would take longer to
get to this website than others. Today for example, I took a few hours to
get to my internet, as I had other activities to cater for.

I just wanted to give you my opinion on your interesting system.

Kind Regards,

Rafael Moraes

 What do u think about it?

I guess the bounce depends on whether the collission is elastic, doesn't it? And I never said it would be instantaneously zero. Or I might have.... oh well, sorry if you took it that way.

The space vehicle may/may not reach the velocity after impacting the lunar surface. It depends on several other factors.

Uh... what kind of cases do we have to give?

 Do we have to state any random formula which requires a dot product?

 1. Dot-product: Work = F.s, where F - force applied, and s - dsplacement of body.

2. Vector-product: Torque about an axis = Force x radius of the circle

 Again, I might have deeply misunderstood the question. And the answer might be dead wrong.

i cant understand this method at all....how r we suppose to know how many points have we earned n how do we get to knw how many marks are we being awarded....

if this method is being taught then please make us understand how will it work and what exactly you want from us... 

Sheet 1

1) A+B:  (1,1,0) + (-1,1,0) = (0,2,0)

2) A-B: (1,1,0) - (-1,1,0) = (2,0,0)

3) 2A+3B: (2,2,0) + (-3,3,0) = (-1,5,0)

4) A.B: (1*-1) + (1*1) + (0*0) = 0

5) AxB:    i      j      k

              1   1   0

             -1   1   0

gives  2k

Sheet 2

S= 5m  U= 2ms-1  V=?  a=9.81/6 => 1.635 T= ? 

V2 = u2+2as

v2 = 22+(2*1.635*5)

v2=407/20 => v = 4.511ms-1

Q1.

A+B = 0, 1, 0

A-B = 2, 0 , 0

 2A + 3B = -1, 5, 0

A.B = 0

AxB = -1, 1, 0

Q2. using v^2 = u^2 + 2as

v being 4.511 m/s

3) If F=ma is applied to a body, will still work, as long as the body is assumed to be a collection of particles, you just have to take all the vectors to the centre of the objects mass.....F is force and a is acceleration

just realised that when i posted my answers before they didnt appear

A+B : (0,2,0)

A-B : (2,0,0)

2A+3B : (-1,5,0)

A.B : 0

AxB : (0,0,2)

2. using SuVat you get 4.511m/s

V^2=u^2 +2aS

Sheet 1

1. (1,1,0) + (-1,1,0) = (0,2,0)
2. (1,1,0) - (-1,1,0) = (2,0,0)
3. 2(1,1,0) + 3(-1,1,0) = (2,2,0) + (-3,3,0) = (-1,5,0)
4. A.B = 1 x -1 + 1 x 1 + 0 x 0 = 0
5. AxB = (1x0 - 0x1, 0x-1 - 1x0 , 1x1 - 1x-1) = (0,0,2)

Sheet 2

v2 = u2 + 2as

u = 2

a = 9.81/6 = 1.635

s = 5

v2 = (2)2 + 2x1.635x5 = 20.35

v = 4.511 m/s = 5 m/s (1 s.f.)

With regards to motion after this, we would need to model the shock absorption as a lunar module will have some sort of system in place to reduce the damage from impact to the rigid structure as much as possible. This could make an interesting extended question if you could give us a model to work with. 

tadaaaa!
q1)
1 - (0,2,0)
2 - (2,0,0)
3 - (-1,5,0)
4 - 0
5 - (0,0,2)

ok, let me do question 2 now... that's all for now!

question 2!

v2=u2+2as

v2=22+2(10/6)(5)

v=4.55m/s

yea?

Cross products are needed for things such as motor effect using vectors.

10 points? 

Thought I might aswell join the club and put my answers up.

Part 1 

Q1) = (0,2,0)

Q2) = (2,0,0)

Q3) = (-1,5,0)

Q4) = 0

Q5) = (0,0,2)

 Part 2

u = 2m/s

s = 5m

a = 1/6*9.81m/s2

Equation  v2=u2+2as

v2=22+(2*5*(1/6*9.81))

v2=20.35m/s

v= 4.51m/s

 Job done

if dot product is zero, the vectors are perpendicular. There is also a formula involving the dot product which allows u to find the angle between the two vectors.

cross product is used to find the normal of the two vectors, this result can help u find the area of a triangle, the angle between vectors, the resulting vector, and many other things...it all depends on what you are meant to find out.

Q1. (1,1,0)

Q2. (-1,1,0)   

Q3. (0,2,0)

Q4. (-1,5,0)

Q5. (0,0,2)

Part 2

v^2=u^2+2as

     =2^2+2x(9.8/6)x5

    v=4.5ms^-1

Im guessing that if Newtons laws are applied to a body, the answer will be much more complex becuase unfortunately unlike particles the mass is not concentrated on one point so the force on different areas of the body will be different so moments will start to become clear.

 If dot product fo vectors=o then they are mutually perpendicular, to expand on Raf's comment a.b=|a||b|cos@  (thats supposed to be a theta btw)

(Q1) A+B = (0,2,0)

 A-B = (2,0,0)

2A+3B = (-1,5,0)

 A.B = 0 AxB = (0,0,2) or 2K

(Q2) h=5m

Initial Velocity= 2m/s

 acceleration= 9.81/6 m/s^2

 We have to find Impact velocity or Final velocity Using the formula v^2-u^2=2as

where v=final velocity u=Initial velocity a=Acceleration s=Distance

so. v^2-2^2=2*(9.81/6)*5

v^2=20.35

 v= 4.51m/s

I am sorry for replying late, as I didnt knew how to use Imechanica.org I had a view about this way of teaching. I know it is good, but it is not a fair way of evaluating the capabilities of a student. We never know that anyone who has posted the answers are fairly done by him, as anyone can copy my work make a few changes in it and post it. What so every you are my teacher, if this is the way you like to teach us then um more then happy atleast i would not have to give my exams :-D