ideatoprince18: pls giv it to Buhari....shebi he said he can solve all our problem

catalanscrew: This is the answer.........................




www.nairaland.com/attachments/2617207_quest_jpeg_jpegb8a74e0b1799acde4add2471d6d75774



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Teempakguy:
this is incorrect. We are asked to find the amount of time spent in the air After the height of 50m has being transversed. I think the best approach would be to find the final velocity at the foot of the jump and use it to calculate the time used in flight.

mickyeddi:

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.

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.

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-(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

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.

Donmichaelz:
the people I know that would give it a shoot has less Dan 100k in their account....All am saying is mathematical problems are not real life problems ,I am more of the latter,dangote can't do these ,neither can I ,,,,

NaijaHotGuy:
Ni wahala o, here is the solution to 4

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-(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

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.

NaijaHotGuy:


Impressive. I love ur analytical style. You really surprised us guys on here. Especially the men. I meant the men anyways. I hate to burst ur burbles dear but i have found three major flaws in ur solution method that led to the wrong answer - I have had to check to see if i cud arrive at the same answer, 3.795.
Yes i have some major related achievements in the past.
The problem is actually about "two steps" more complex.
Beyond the maths it will really be great to interact with you some day
Would also want to know if u cud solve some problems i cud solved in SSS 3 (way long time a ago)
I will be needing some indication of interest here from you before uplading my rough work but with explanation

Dollyak:

I love being challenged so you can upload your work

Dollyak:

I love being challenged so you can upload your work