How many golf balls can fit in a school bus?

Depends how big the school bus is, for simplicity, lets say its the small bus that can fit 20 + 1 people. You have to take account of the empty space as well, therefore a person could be sitting in the middle line and three more in the front so we have around 30 people making a full squished bus. The average size of a golf ball is around 3 cm in diameter. The height of the bus is usually 200 cm, the width of an average person is usually 50 cm. The depth of the seat is about 70 cm You can calculate the volume of that cubic form to be 50 x 70 x 200 x 3 x 30 = the number of golf balls in average.

Nobody need to actually put golf balls in a school buss , nobody care if the number is 61,000,000 or whatever .Nobody in Google cares how good are you in giving estimates to the size of a buss. The answer is bussWidth*bussHight*bussLength / (ball/Diameter ^ 3 * sin (60) * sin(60) ^2 ) . This answer depends on the size of the school buss, different busses yield different answer. The sin(60) are since its gold balls and not cubics.

As Jehova put it. This tests your problem-solving stance being seeing that you leave the right things as variables. The question is really about trigonometry and understanding how tightly you can pack spheres. In other words, you need to figure the effective volume of each golf ball. To do so, first consider a single plane of balls, and a single row from it. If the ball radius is r, the width required for each ball is 2r. Now consider the depth required by each row. Since the rows will be "staved off" from each other, each row requires somewhat less than 2r for its depth. In fact, each requires r * tan 60, or about 1.73 * r. To see why, notice that each adjacent group of three spheres forms a triangle, and, because of the symmetry of the situation, their centers form an equilateral triangle. Thus each of its angles it 60 degrees. Since tan(theta) = opposite / adjacent, we have that tan(60) = depth / radius, or depth = radius * tan(60). Finally, consider the height require by each plane of balls. Take the three adjacent balls from the first plane and imagine adding a forth ball above them in the center. The centers of the four balls now form a tetrahedron (just a pyramid with a triangular base). We need to find the height from the center of the base of the tetrahedron to its top vertex. Looking at a diagram, one can see that the distance to the center of the base from one of the base vertices is found by cos(theta) = radius / hypotenuse, or cos(30) = r / distance, meaning distance = r / cos(30), or about 1.15 * r. If you create a line going from the base vertex to the base center, it then becomes clear that this line forms a vertical isosceles right triangle with the height line. In other words, the distance we just found is equal to the height of the tetrahedron. Bring it all together, we have that the effective volume of each ball is: effectiveWidth * effectiveDepth * effectiveHeight = 2r * r tan 60 * r / cos 30. So finally, the number of balls is the bus volume divided by each ball's effective volume.

"If you create a line going from the base vertex to the base center, it then becomes clear that this line forms a vertical isosceles right triangle with the height line. In other words, the distance we just found is equal to the height of the tetrahedron." My girlfriend actually figured it out better when I got home. I was wrong saying it was clear that the internal vertical triangle was isosceles. It's not actually isosceles. The way to figure out the height of the tetrahedrom is actually to note that the hypotenuse is 2r and use the Pathagorean(sp?) Theorem: height^2 = (2r)^2 + (r / cos 30)^2, and solve for height. Sorry!

not enough to shove up your a@@

@Anonymous commenter on 9/10/2011: you are completely ignoring the effective volume inside a bus. You would have to subtract volume taken up by seats as well as any people inside the bus.

http://mathworld.wolfram.com/SpherePacking.html

48 passenger + 1 driver therefore 49 * 2 = 98

A school bus of golf balls.