Can some one give me at least 3 difficult ratio question?

Question

i want 3 maths ratio question that is 'very' hard.pls put the question and solution. pls submit it by 1st september. thanks for your help .and i want ratio . ratio!!!

Answer 1

1.  A pitcher can throw a ball at 100kph (regardless of the angle he throws it at).  To break a window, the ball must be traveling at least 55kph in the horizontal direction.  

a.  Why can't we conclude he will always be able break a window?

Answer:  If the window is very high off the ground, most of the 100kph is used to raise the ball to the correct height and much less of it will be available for horizontal speed.  Example, if he throws the ball directly vertically there is no horizontal speed and it couldn't possibly break the window.

b.  Recall that throwing ball at an angle of 45 degrees above the ground will enable the ball to travel the furthest distance.

If the pitcher must throw the ball at a 45 degree angle to reach the window, will the ball break the window.  (hint: use the pythagorean theroem)

Hypoteneus = 100kph

Side A = Side B (property of a 45, 45, 90 triangle)

a^2 +b^2 = 100^2

2a^2=10,000

a = 5,000^(0.5)

70.71kph in the horizontal  

Yes, the window will break!

c.  At a 45 degree angle, what proportion of the inital speed translates into the horizontal direction?  Into vertical direction?

(70.71/100) , (70.71/100)

d.  Using the proportion found in part c, what is the slowest pitch thrown at 45 degrees that will break the window?

(55/0.7071) = 77.78 kph

2.  We have an ordinary roll of toilet paper that is rolled down a hill such that it unravels.  The thickness of a sheet of toilet paper is 1 milimeter, the diameter of the hole in the middle is 4mm, and the total diameter of the roll is 20mm.  Find the appoximate radius of the roll when exactly 1/3 of the paper has unravelled.

Easier than you think.  Each time one layer unravels, the radius decreases by 1mm and 2(pie)r millimeters of paper unravel.  

Initial radius 10mm, and this layer unravels a distance of

(2)(pie)(10)mm or 62.8mm.  The next layer has radius 9mm and unravels (2)(pie)(9)mm or 62.8mm + 56.52mm and so on until the radius is 2mm and no paper unravels.

(2 * 3.141)(10 + 9 + 8 + 7 +6 + 5 +4 +3)   This gives the total length of the paper on the roll.

Total Length= 326.56mm

1/3 unraveled at 108.85mm

After the second layer has completely unravelled, the roll has already traveled 62.8 + 56.52 = 119.32mm; slightly greater than 1/3 of the total roll so the radius must be slightly bigger than 8mm.

Lets find the approximate radius  by finding the ratio of the second layer that has unravelled.

108.85 - 62.8 = 46.05mm contributed by sceond layer

(46.05/56.52) = 0.815 or 81.5% of the second layer has unravelled.

1 - 0.815 =.185 so the radius is 8.185mm on average.

I'll come up with something else later, is this too difficult?

Answer 2

What ratio is the length of rope on a tethered sheep to the diameter of paddock, so it can eat half an acre in 1 acre round paddock