a dartboard has a diameter of 19inches the bullseye's diameter is 1/2in?
353709
what's the "beer" variable?
436542
There was a English golf ball called Robinhood that was smaller than the American ball.  Sometimes you find them on e-bay
1013311
according to the International Tennis Federation the official diameter as "more than" 2 and1/2 inches but "less than" 2 and 5/8 inches. The weight is "more than" 2 ounces but "less than" 2 and 1/16th ounces
81906
First Question:\nVertical and horizontal motion are independent from each other.  Use d=(v.i)(t)+(1/2)(a)(t)^2 to find the time that the dart is in the air.  d is distance(0.32m), v.i is the initial velocity(0.00m/s vertically since it was thrown exactly horizontally), a is acceleration (9.8m/s^2: gravity) and t is the time in seconds.  Now that you have the time, use that in the equation x=vt, where x is the horizontal distance (range) that you are asked to solve for, v is the velocity(11.8m/s), and t is the time that you just found.  The answer to the problem is the x variable in this equation.\n\nSecond Question:\nI have attempted this problem for about 40 minutes and I have tried every possible combination of variables and equations, but there is not enough information to solve it.  Therefore, I have to say that it is an impossible problem.  If you figure out how to solve it, please, send me an e-mail at the address I give later and tell me how to do it.\n\nThird Question:\nFirst, find the distance that the basket is from the launch point by subtracting the two.  Then, use (v.f)^2=(v.i)^2+(2)(a)(d), where v.f is the final velocity, v.i is the initial velocity, a is the acceleration (9.8m/s^2), and d is the distance which you just found.  To  get the ball to the basket, it only needs to go the exact distance to the basket, so the final velocity is 0.00m/s.  Use the previous equation to find the intial vertical velocity.  The use v.iy=(v)sin(angle), where v.iy is the initial vertical velocity and v is the initial angular velocity, to find the velocity the ball must be launched at for each angle.\n\nPlease send me an e-mail at bge1234@yahoo.com after you turn in your work and get the solution so that I can see how your teacher did it.  In particular, tell me how to do the second question.  My ways SHOULD work as long as you don't make any calculatory errors.  I hope this helps you.
176943
2 inches? maybe?  I don't know, one so small that it can't penetrate.
811862
(8/12)(1/4)\n=1/6 inch
870436
About 259 miles.  This is based on the diameter of a BB = 0.17 in.
886387
half an inch---- 0.5 in.
615678
1 inch= 2.54 cm\n\nequivalently...\n\n1 inch= 25.4 mm
793793
the thing this problem is concerned is the vertical component of the projectile.\n\nsince the dart is thrown horizontally, the initial vertical velocity is zero.\n\nthe vertical velocity of the dart just before it hits the dart board is\n\nv^2 = u^2 + 2gs\nv^2 = 0 + 2(9.8)(0.32)\nv^2 = 6.272\nv = 2.50 m/s\n\nthe time it took the dart to hit the dart board is given by\n\nv = u + gt\n2.50 = 0 + 9.8t\nt = 0.255 sec\n\n\nnow, we have the time of travel of the dart, we can use the formula  d = vt  to find the horizontal distance\n(the horizontal velocity remained constant throughout the flight)\n\nd = vt\nd = 12.4m/s * 0.255s\nd = 3.169 m\n\n\ntherefore the dart player is standing  3.169 meters away from the board...\n\nthere ya go...
740272
1/4"= 1'\n1" =4'\n12" or 1' = 4*12=48' Ans.
909088
2.54 cm is the correct answer to that one
14438
Your board will need to be hung exactly 5 feet 8 inches from the floor to the center of the bullseye.  Keep in mind, if a backboard is used to protect the wall around the board, measurements should be considered to keep the backboard centered around the dart board.\n\nMarking Your Throw Line\n\nA toe line is either a raised oche or a length of tape placed along the floor parallel to the wall the dart board is mounted on.  The throw line must be a minimum length of (24").  Furthermore, a raised oche is specified and must be a minimum of (1.5") high.  Whichever type of throw line you decide upon, it should be placed - \n\nExactly 7 feet 9 1/4 inches from the face of the board for STEEL-TIPPED DARTING.\n\nor\n\nExactly 8 feet from the face of the board for SOFT-TIPPED DARTING.
971117
So 3/6 is in sand and 2/6 is in water.  So 1/6 is above water and thats 1".\n\nTherefore the stick is 6" long
127476
25 by 14 inches, which is 2ft. and 1 in. by 1ft. and 2in.
893271
1 inch = 2.54 cm\n\nSo 1 cm = 1/2.54 inches \n\nSo n cm = n/2.54 inches
237156
Pi * r^2\n\n(Pi R Squared)\n\nwhere R is the radius (which is also equivalent to half the diameter)