Dollyak:
Number 2. Attempts. The second easiest after 4

Babe well done! but the pic no clear oh

Adultnextdoor:
......it doesn't matter whether he is the op or not......his calculation is incomplete but its obvious he has an idea!

Hence you are so good in maths, can you solve question three so that we know your level? Or give us theoretical background answer to the question.
Bear in mind I have just solved it.

I need a refresher course, can't remember how to do any of these.

Adultnextdoor:
......now.... This is a complete and correct calculation.......i'll give you 10/10....

Just an explanation, thus it's not even complete. Shows you don't even know what you are talking about.

Adultnextdoor:
.......i'll give you a laid down calculation very soon.

Waiting....

Starships4u:
You chop craze? Y I go solve am Na me spoil am

Dollyak:
Number 4 is quite easy too as well.
If you think about this. The total amount spent in the air by the skier is the same as the time taken to slope down(starting from the horizontal position) the whole distance of 50meters.
thus distance gives =1/2 at t^2
Therefore 50=1/2 × gt^2 = 1/2 (9.8 )t^2 =50 solve for t
T= 3.2 seconds

Very easy because time only depends on gravity in this case.

You didn't make use of the coefficient of friction (0.05)?

No1 answer is 3 since 3 raise to power of 6
Is more than 500 than n can take the value 0 1 or 2 it me hyperbole the best in calculation but also think along with that...! It is

Where is prof ibime...let him come here

If possible I will come back to this life...my request to God bii say make him help me build ma brain reach level wey I go fit solve any mathematics...cus if I see snake nd math I go run for math...

Cc:
Agentofchange1
frob0genius
Dejt4u

Adultnextdoor:
......you tried but not there yet.......i'll give you 4.5/10 for the attempt..

guy dat great work that what I also did hyperbole u are actually correct it is just applied mathematics when you need only you common sense

writetopoker:
as I see dis maths my mind come fly, but I catch am bck, op u wan make I craze abi? as I see dis maths my mind come fly, but I catch am bck, op u wan make I craze abi?

from my observations its like u are crazy already

pls giv it to Buhari....shebi he said he can solve all our problem

destinysaid:
Maths ke, give me chemistry and see me work magic

so u think u kno chemistry uhn...?
i laff!

joywendy:
Cc:
Agentofchange1
frob0genius
Dejt4u

yesss my lady ..I heard my name from afar .....

don't mind that dude ..those are Olympiad questions. .....would av solve them ..but 2 Prove what exactly,?. I know what I know , don't nid 2 prove my smartness 2 any nigga ... I only solve questions to help others. & 4 fun..


hope u get ma gist .?..

Badgers14:

Baba na u o

eskober:
from my observations its like I am crazy already

may your soul rest in pieces, bloody idiat!

stuffs like this make me sick

This is projectiles.my try
max.height=u2sin2tita/2g
max.height=5m
5=u2sin2(30)/20
100=u2x0.25
25=u2
u=5m/s
still loading.. . . .

Am coming. Wait for me, make i get my market calculator.

If you judge a fish by it ability to climb a tree, it will live the rest of its life thinking you are a fool. Have u heard of that saying before? Though I can proffer solution to those questions.

This is the answer.........................








.......

immediately i read the word "integer"...i just scrolled down. no need..

Obinoscopy:
You didn't make use of the coefficient of friction (0.05)?

Factoring the coeeficient of friction mean the net force on the skier on the ski jump parallel to the ski jump surface is a combination of the gravitation force and the frictional force,
For simplicity
F(net)= mg (sinθ -μcosθ)

The net work done on the skier as he travels down the jump is W= FL where L is the length of the jump, L=50/sinθ


By the work-energy theorem the skier launches horizontally from the bottom of the jump with a speed
v(=(2gh(1-μcotθ))^2
After the skier launches (and treating the launch point as x,y=0), the horizontal and vertical position of the skier are
and at the moment of impact, (x,y)=(0,0)
x=vt, y=-1/2 (gt)
The landing spot is on the line , so we have a system of equations we can solve for t=3.794 seconds.
v=29.919m/s check attached pictures to see how I solved for v


More detailed calculation using conservation energy. ..

Using the conservation of energy principle, the work done by friction must be equal to the change in the mechanical energy of the system (h = 50m, alpha = 30°),
(1/2mv^2)-mgh=-μmgcosθ(h/sin30)
This implies
v=sqrt (2gh (1-μcot30))
During the fall, using parallel (x axis pointing 'downwards') and perpendicular (y axis pointing 'upwards') axes in respect to the hill, we can write down the variation of y with time as
y=y[size=4pt]0[/size]+vsin30t-(g cos at^2)/2)
Setting y=0 and solving for t(h'=5m)
t=vtana/g (1+sqrt (1+(2gh'/v^2 tana^2)))=3.795

T is 11.9 Seconds

mickyeddi:
They are ranked by a site called Brilliant.org, which asks users progressively harder math and science problems.

Think you might be as smart as one of these tweens or teens?

Prove it by solving some of the math problems they've solved.






1.


2.


3.


4.

solla22:
T is 11.9 Seconds

t is not 11.9. How did you solve it.

Dollyak:

Factoring the coeeficient of friction mean the net force on the skier on the ski jump parallel to the ski jump surface is a combination of the gravitation force and the frictional force,
For simplicity
F(net)= mg (sinθ -μcosθ)

The net work done on the skier as he travels down the jump is W= FL where L is the length of the jump, L=50/sinθ


By the work-energy theorem the skier launches horizontally from the bottom of the jump with a speed
v(=(2gh(1-μcotθ))^2
After the skier launches (and treating the launch point as x,y=0), the horizontal and vertical position of the skier are
and at the moment of impact, (x,y)=(0,0)
x=vt, y=-1/2 (gt)
The landing spot is on the line , so we have a system of equations we can solve for t=3.794 seconds.
v=29.919m/s check attached pictures to see how I solved for v


More detailed calculation using conservation energy. ..

Using the conservation of energy principle, the work done by friction must be equal to the change in the mechanical energy of the system (h = 50m, alpha = 30°),
(1/2mv^2)-mgh=-μmgcosθ(h/sin30)
This implies
v=sqrt (2gh (1-μcot30))
During the fall, using parallel (x axis pointing 'downwards') and perpendicular (y axis pointing 'upwards') axes in respect to the hill, we can write down the variation of y with time as
y=y[size=4pt]0[/size]+vsin30t-(gcosat^2)/2)
Setting y=0 and solving for t(h'=5m)
t=vtana/g (1+sqrt (1+(2gh'/v^2tana^2)))=3.795

aunty,pls i cant understand this oo...use x as your variable instead of a..thanks

Adiwana:

aunty,pls i cant understand this oo...use x as your variable instead of a..thanks

a is the alpha=30 degrees. It is very hard to type equations using phone. Can't be bothered to use Microsoft Word lol

Adiwana:

aunty,pls i cant understand this oo...use x as your variable instead of a..thanks

hope you understand?

Dollyak:

hope you understand?

nawao

Cc: Teempakguy
Johnydon22
Robinhez

Come solve this maths