A C D die

The stone pushed by 'E' is not likely to bounce and kill 'b' because, even though it appears lighter than the other stone, the momentum it will gather from rolling down the slope will make it lift the other stone a little, balance and drop it back. My take.

solid3:
Take a look at this pix and tell us who dies with reason.


*solid*

I would love to but first, wetin carry them go dia. Children listen to you parent.

Final Destination 10...

Well realistically, it's likely we wouldn't have the exact same outcome if that scene plays itself out twice. When you cut out a part of a stone, a portion as large as that, it would not roll perfectly, even down a slope. Maybe a combination of sliding and rolling. Only the perfect roll would save D's head.

If D doesn't die, it's possible no one else would. It's also possible everyone else would. Depends mostly on the force Mr E applies with his push.

Martiyu:


do u know dt v been on d lookout for u?...OK...now m watching u.

Huh

Baddestguyp:
all except b. only a Sharp mind will notice how A died

Please explain.

B is dead. No argument

Everybody dies. Cos even if they don't die in this puzzle they'll sure die in another one you post.

Let's call L the length of the slope : L = sqrt(4^2
+ 3^2) = 5 squares
R is the ball's radius : R = 1 square
The perimeter : P = 2*Pi*R = 6.28 squares
The "percentage of rolled ball on the slope" :
T = L/P = 0.796 = 79.6%
So the rolled angle A : A = T*360 = 286
degrees
Then, as the hole on the ball is hard to mesure with
squares, I used pixels. It's 25 pixels large, when 7
squares are 255 pixels large.
I got l the hole size (at the top) :
l = (7*25)/255 = 0.686 squares
So : the hole angle equals the angle in the top corner of
the isosceles triangle, wich have as base : the hole, and
as top corner the center of the ball. Trigonometry :
tan(a/2) = (l/2)/R
a/2 = arctan((l/2)/R)
= 0.33 radians
a/2 = 19 degrees
Finnaly :
A-a/2 = 267
A+a/2 = 305 degrees
and 180 < 267 < 305 < 360+180
"half turn" < "rolled angle +/- hole size" < "one and a half"
When the ball is above the D-man, it doesn't strand into
the hole. So D survives.

hmmmm....na E, C, and D

Where is d op to settle dis but I think C will die

Codedboy95:
Let's call L the length of the slope : L = sqrt(4^2
+ 3^2) = 5 squares
R is the ball's radius : R = 1 square
The perimeter : P = 2*Pi*R = 6.28 squares
The "percentage of rolled ball on the slope" :
T = L/P = 0.796 = 79.6%
So the rolled angle A : A = T*360 = 286
degrees
Then, as the hole on the ball is hard to mesure with
squares, I used pixels. It's 25 pixels large, when 7
squares are 255 pixels large.
I got l the hole size (at the top) :
l = (7*25)/255 = 0.686 squares
So : the hole angle equals the angle in the top corner of
the isosceles triangle, wich have as base : the hole, and
as top corner the center of the ball. Trigonometry :
tan(a/2) = (l/2)/R
a/2 = arctan((l/2)/R)
= 0.33 radians
a/2 = 19 degrees
Finnaly :
A-a/2 = 267
A+a/2 = 305 degrees
and 180 < 267 < 305 < 360+180
"half turn" < "rolled angle +/- hole size" < "one and a half"
When the ball is above the D-man, it doesn't strand into
the hole. So D survives.

wat in the world is this?

Codedboy95:
Let's call L the length of the slope : L = sqrt(4^2
+ 3^2) = 5 squares
R is the ball's radius : R = 1 square
The perimeter : P = 2*Pi*R = 6.28 squares
The "percentage of rolled ball on the slope" :
T = L/P = 0.796 = 79.6%
So the rolled angle A : A = T*360 = 286
degrees
Then, as the hole on the ball is hard to mesure with
squares, I used pixels. It's 25 pixels large, when 7
squares are 255 pixels large.
I got l the hole size (at the top) :
l = (7*25)/255 = 0.686 squares
So : the hole angle equals the angle in the top corner of
the isosceles triangle, wich have as base : the hole, and
as top corner the center of the ball. Trigonometry :
tan(a/2) = (l/2)/R
a/2 = arctan((l/2)/R)
= 0.33 radians
a/2 = 19 degrees
Finnaly :
A-a/2 = 267
A+a/2 = 305 degrees
and 180 < 267 < 305 < 360+180
"half turn" < "rolled angle +/- hole size" < "one and a half"
When the ball is above the D-man, it doesn't strand into
the hole. So D survives.

I don laff tire u funny die wit all these jargons

Nwachiizu:


Please explain.

made a mistake, only d , c and A will die.
the rock cuts off D's head and due to d speed at which it lands, it goes down for a second killing C and the weight of d heavier rock throws it up to innocent A

post=27672297:
They all die

-------------
Fellow Nairalanders, I have been a member of this forum for more than 8 years and one way I can benefit from it is for you guys to help patronize my blog at yemisisblogdotcom ...Abeg, I have no where to run to
-------------

your blog link is not opening...... both in your profile and on. your signature.......I tried clicking on it but no response .....so edit it.....

Codedboy95:
Let's call L the length of the slope : L = sqrt(4^2
+ 3^2) = 5 squares
R is the ball's radius : R = 1 square
The perimeter : P = 2*Pi*R = 6.28 squares
The "percentage of rolled ball on the slope" :
T = L/P = 0.796 = 79.6%
So the rolled angle A : A = T*360 = 286
degrees
Then, as the hole on the ball is hard to mesure with
squares, I used pixels. It's 25 pixels large, when 7
squares are 255 pixels large.
I got l the hole size (at the top) :
l = (7*25)/255 = 0.686 squares
So : the hole angle equals the angle in the top corner of
the isosceles triangle, wich have as base : the hole, and
as top corner the center of the ball. Trigonometry :
tan(a/2) = (l/2)/R
a/2 = arctan((l/2)/R)
= 0.33 radians
a/2 = 19 degrees
Finnaly :
A-a/2 = 267
A+a/2 = 305 degrees
and 180 < 267 < 305 < 360+180
"half turn" < "rolled angle +/- hole size" < "one and a half"
When the ball is above the D-man, it doesn't strand into
the hole. So D survives.

hm! Na wa o

Codedboy95:
Let's call L the length of the slope : L = sqrt(4^2
+ 3^2) = 5 squares
R is the ball's radius : R = 1 square
The perimeter : P = 2*Pi*R = 6.28 squares
The "percentage of rolled ball on the slope" :
T = L/P = 0.796 = 79.6%
So the rolled angle A : A = T*360 = 286
degrees
Then, as the hole on the ball is hard to mesure with
squares, I used pixels. It's 25 pixels large, when 7
squares are 255 pixels large.
I got l the hole size (at the top) :
l = (7*25)/255 = 0.686 squares
So : the hole angle equals the angle in the top corner of
the isosceles triangle, wich have as base : the hole, and
as top corner the center of the ball. Trigonometry :
tan(a/2) = (l/2)/R
a/2 = arctan((l/2)/R)
= 0.33 radians
a/2 = 19 degrees
Finnaly :
A-a/2 = 267
A+a/2 = 305 degrees
and 180 < 267 < 305 < 360+180
"half turn" < "rolled angle +/- hole size" < "one and a half"
When the ball is above the D-man, it doesn't strand into
the hole. So D survives.

see mathematics. abeg which formula be this

The stone being pushed is lighter than the stone being blocked from B as a result of the part cut out. The cut-out part make execution of D impossible as it roll down, C is safe because the other stone is heavier so the lighter stone isn't able to press the spear ends on C as it throts forward due to the part of the plane that inclinates before the heavier stone, thus projecting it forth to land on B. Shall we please all observe a minute of silence for the unfortunate one.

davroca16:
so then we'll settle for 50-50 survival for D.

He has less than 50,
Look again cos the part of stone chop off isn't up to 40%

solid3:
Take a look at this pix and tell us who dies with reason.


*solid*

As the chopped stone rolls down from E, D is having a possibility to survive if it is the chopped part that passes through his head, but from the analysis of the graduation made, death is sure for D as the chopped part will only be facing the ground for just once before getting to D. Due to the fact that the chopped stone will have its weight lesser than that of a complete stone, as it moves away from B, the complete stone will prevent it from hitting C and as such move towards the complete stone and tilt a bit upward, flying to fall on B who is the only person with full assurance to die.

Codedboy95:
Let's call L the length of the slope : L = sqrt(4^2
+ 3^2) = 5 squares
R is the ball's radius : R = 1 square
The perimeter : P = 2*Pi*R = 6.28 squares
The "percentage of rolled ball on the slope" :
T = L/P = 0.796 = 79.6%
So the rolled angle A : A = T*360 = 286
degrees
Then, as the hole on the ball is hard to mesure with
squares, I used pixels. It's 25 pixels large, when 7
squares are 255 pixels large.
I got l the hole size (at the top) :
l = (7*25)/255 = 0.686 squares
So : the hole angle equals the angle in the top corner of
the isosceles triangle, wich have as base : the hole, and
as top corner the center of the ball. Trigonometry :
tan(a/2) = (l/2)/R
a/2 = arctan((l/2)/R)
= 0.33 radians
a/2 = 19 degrees
Finnaly :
A-a/2 = 267
A+a/2 = 305 degrees
and 180 < 267 < 305 < 360+180
"half turn" < "rolled angle +/- hole size" < "one and a half"
When the ball is above the D-man, it doesn't strand into
the hole. So D survives.

According to your calculations, what percentage of the stone is chopped off? Or likely percent

erad:
Nobody dies...

The curved out part on the upper ball makes it lighter than the ball on the other end of the scale. So it'll miss the head of the standing guy and won't have enough weight to lift the other stone. It'll end up rolling to the other side where the complete stone is.

My social scientific opinion.

Rubbish.

Only E is safe......

A, B, c, and D dies
Stop complicating simple stuff with analysis

but nairalanders can discuss and analyse anything sha ? even crap



choi!!!

post=27672297:
They all die

-------------
Fellow Nairalanders, I have been a member of this forum for more than 8 years and one way I can benefit from it is for you guys to help patronize my blog at yemisisblogdotcom ...Abeg, I have no where to run to
-------------

says who?abeg run to my arms

None will die.......cos they r all safe.

[img][/img]http://www.funnyjunk.com/Who+dies+puzzle/funny-gifs/5341271/ the first and the last guy both survive

Codedboy95:
Let's call L the length of the slope : L = sqrt(4^2
+ 3^2) = 5 squares
R is the ball's radius : R = 1 square
The perimeter : P = 2*Pi*R = 6.28 squares
The "percentage of rolled ball on the slope" :
T = L/P = 0.796 = 79.6%
So the rolled angle A : A = T*360 = 286
degrees
Then, as the hole on the ball is hard to mesure with
squares, I used pixels. It's 25 pixels large, when 7
squares are 255 pixels large.
I got l the hole size (at the top) :
l = (7*25)/255 = 0.686 squares
So : the hole angle equals the angle in the top corner of
the isosceles triangle, wich have as base : the hole, and
as top corner the center of the ball. Trigonometry :
tan(a/2) = (l/2)/R
a/2 = arctan((l/2)/R)
= 0.33 radians
a/2 = 19 degrees
Finnaly :
A-a/2 = 267
A+a/2 = 305 degrees
and 180 < 267 < 305 < 360+180
"half turn" < "rolled angle +/- hole size" < "one and a half"
When the ball is above the D-man, it doesn't strand into
the hole. So D survives.

all these because of this simple question?i must report you to seun for wasting his land.

0Gimme the density/weight of both balls and materials they are made of, the initial velocity of the ball, the force exerted by E on the ball, what part of the planet dey r in so I cld know what free fall gravity to imply, if D is a dummie (coz I expect it to dodge a rolling stone by impulse)... Provide me with these answers, then I'd tell you who dies. That if there'll be any.

The most likely to die is D follow by C while E takes off, A and B will snap pic of either D or C to upload on nairaland.