₦airaland Forum

See this one

The funny thing is that , I was replied that my aggregate was the lowest in the group so far that 68 was the lowest before

TeflonBlixx:
Lol. . Did you ask them to post screenshot?

Most of them lie about their results

I never asked oo

Pes13:
If they are not lieing, then the cutoff for mech can't be less than 70 definitely.

Thanks boss for your reply
I thought as much

Pes13:
If they are not lieing, then the cutoff for mech can't be less than 70 definitely.

Thanks boss for your reply I thought as much

simplebb:
That's serious. As aggregate score?

Yeah Aggregate score.
When I joined the WhatsApp group I ask for the highest so far and someone replied he can't count that they are up to 40 in the group.

scromatan:


Like serious??

Very serious

TeflonBlixx:
Guy.. you see those high marks Kai!! Prepare your mind o

I joined a mech eng. WhatsApp group yday omo about 40 of them had 70 above. I wanna faint

scromatan:
no sire


Mathematics

you??

Mech eng.
And really bothering about cutoff mark

TeflonBlixx:
75.375

Agba

scromatan:
.....

....

...

..

.

Engineering aspirant?

Pes13:
According to the school calendar, we are finishing this session December, a month break, then we resume February.

You guyz are expected to resume a week before us.

All admission process including sch fees and hostel fee would have been done with and dusted before February.

Please boss what can you say about mech eng aspirants, what the cut off mark may likely be.
My aggregate is low 63.75 and last year was 64.375

Please o guys, any idea of what the cut off mark for mech eng. Would be. Because my aggregate is quite low.

Mechanical engineering student drop your scores
Mine
63.75 good to go?

[quote author=overcomer90 post=70043778]Distance between top of tower and maximum ht. is given by V²= U² + 2as
U = initial velocity = 20m/s
At maximum ht. V = 0
a = -10m/s since it is initially moving upwards
We have that
0= 20² + 2*-10*s
20²= -20s
=> s = 20m (This is the distance between top of tower and maximum ht.)The time taken for this initial motion is given by s = ut + 1/2at²
Where
s is now = 20m
u = 20m/s
t =?
a=-10m/s²
Solving for t u have that t = 2 secs (time taken to reach maximum ht.)

The next step u find the distance from the maximum ht to d ground given by s = ut + 1/2at²
The time here is the total time - the time taken to reach maximum ht.
= 6-2
t= 4secs
U = 0 when it is falling down
a= 10m/s²
Solving for s u have that
s = 80m ( distance from the maximum ht to d ground)

Last step
Height of tower =(distance from the maximum ht to d ground - distance from top of tower to the maximum ht)
H = 80m - 20m
H = 60m

[/quote/]
correct

A small metal ball is thrown vertically upwards from the top of tower with an initial velocity of 20ms-1. If the ball took a total of 6s to reach ground level, determine the height of the tower. ( g = 10ms-2)

Am interested,is it online base or what?

Am interested

Damayogenius18:
A wire is stretch to make it 0.2% longer....what is the percentage change in resistance....(A) 10% (B)0.2% ( C)0.4% (D)5%

B

Is there a WhatsApp group for this thread?
Am Mech engineering aspirant

Please guys I need ui postume past question
Here is my � Quazeemastic@gmail.com

Have anybody seen his or her result (13thof march candidate)

Which info?

Thanks bro i gave you a qall

help me on this i,just got rejected by 2 networks:Maxbounty and G4offers

Please can someone give the steps in sign in up for CPA networks as a beginner without being disapproved.i will be very glad on your opinion

Without changing country wont i be accepted?

I am beginner please help me out on this.i will be very glad for your assisst

LOLZ