if a 6 pound bowling ball is bowled down a 60 ft. alley in 1.7 sec what is the equivelant in a baesball pitch?
466899
Jesus. Can we not use something a bit simpler than these weird pounds, mph etc?\n\nOK, 6 pounds is 2.75 kg, let's call it 3 kg. 60 feet is about 20 metres. it does it in 1.7 seconds. That's 12m/s \n\nSo ball Kinetic Energy is .5mv^2. So .5*3*12*12 = 216 Joules\n\nA baseball weighs about .15 kg. Swapping the formula round:\n\nv= sqrt(2*216/.15)= 60 m/s for the same energy. About 216 kph, or 134mph\n\nI think....
67586
60 feet and six inches is the distance between home plate and the pitcher's mound.\n\nFor everyone else that is 18.4 meters.
60154
http://www.zelo.com/bowling_terms.asp
230718
Answer:\n8 seconds\n\nProof:\nThe problem is really a question of finding what value(s) of x make y equal to zero in the given equation.\nWhen y (the height or distance) equals 0 then the ball is either leaving the ground or hitting the ground. The values of x (the time) that make the height equal to zero tell us when this is happening.\nThe given equation is an inverted parabola so there may be two points where y = 0.\n\n1) Solution 1: x = 0\ny = -16x^2 + 128x\ny = -16*(0^2) + 128*0\ny = -16*0 + 0\ny = 0 + 0\ny = 0\nBut x=0 refers to the time when the ball is first thrown. We want to find the time when the ball lands.\n\n1) Solution 2: x = 8\ny = -16x^2 + 128x\ny = -16(8^2) + 128*(8)\ny = -16*64 + 1,024\ny = -1024 + 1,024\ny = 0\nWhen x (the time) equals 8 seconds, y (the height) equals 0. So ball hits the ground after 8 seconds.\n\nFor your information, after 4 seconds the ball reaches its maximum height of 256 feet.
1150360
The acceleration due to gravity is 32.2 ft. per second per second, which is pretty near 32.  Now we need two fundamental equations of physics:\n\nv = a / t\n\nwhich we can re-write\n\nt = a / v\n\nSince a (acceleration) is 32 ft/sec^2, and v (velocity) is 32 ft/sec, time (t) is 32/32 = 1 second until the ball's velocity equals zero.\n\nSo how much distance in 1 second? The second equation gives us the answer:\n\nd = 1/2 a t^2\n\nwe know a = 32, and since we now know t = 1, we can solve the equation. \n\nt^2 = t * t = 1 * 1 = 1\n\nso\nd = 1/2 * 32 * 1\nd = 16 feet
507693
I'll get you started...\n\nFirst, determine how long the ball was in the air.  From that, determine the velocity at which the ball was thrown (the rate necessary, factoring in gravity's acceleration, to keep it airborne for that period of time).  Construct a vector using the velocity of the truck vs. the velocity of the ball to determine the angle.\n\nGood luck!
671573
Sixty feet, six inches
414794
7-10-1? three pins? that's much easier than just 7-10. take a 8 lb ball, go from left and aim at the right of the 1 pin. let the 1 bounce and hit the 7 while the ball will also bounce to hit the 10! Got it?
756287
Yes, Kaplan is correct.  While the ball is being thrown it is accelerating (not necessarily at 15 m/s2) up to a velocity of 15 m/s.  Once the ball is released the only force acting on it is gravity at -9.8 m/s2.
604691
60 feet 6 inches in professional baseball\n\nlittle league is 46 feet
534824
60mph is 88 ft/sec; (memorize that).\nSo, 90mph is 132 ft/sec.\n\nIgnoring air friction (which DOES have an effect!), it takes 90/132 of a second for the ball to travel that far - or about .682 seconds.\n\nHow far does something drop in about 2/33 of a second? Several feet.
147491
The previous idea of doing an experiment or two is a good idea. Here are some more ideas:\n\nThe ratio of the height you drop the ball from to its rebound height is called its coefficient of restitution. Different types of squash balls have different coefficients of restitution. \n\nThe square root of the coefficient is also the ratio of the maximum speed of descent to maximum speed of rebound (converting potential energy into kinetic energy, and vice versa).
1033468
2.36 m s^-1 to 2 d p\n\nWe split the launch velocity of the ball into a vertical and a horizontal component.\nHorizontal component = 19.8 * cos(49.1) = 12.9639 m/s\nVertical component = 19.8 * sin(49.1) = 14.9659 m/s\n\nGravity accelerates the ball vertically with the acceleration -9.8 m s^-2 (negative because the pull of gravity is downwards while the ball is moving upwards to start with)\n\nThe vertical distance from the ball to the ground after time t is given by the formula s = u * t + (g * t^2) / 2 where: u = initial velocity, g = acceleration of gravity, t = time elapsed\n\nWhen the ball returns to the ground, s = 0\nTherefore the time taken for the ball to return to the ground is given by\nu * t + (g * t^2)/2 = 0 Substitute values\n14.9659 * t - (9.8 / 2) * t^2 = 0 Remove brackets, divide by t, rearrange\n4.9 * t = 14.9659\nt = 3.05427 s\n\nDuring that time the ball will travel horizontally \ns = u * t where s is distance in m, u is velocity in m s^-1, t is time in sec\nDistance travelled horizontally is 12.9639 * 3.05427 = 39.5953 m\n\nThe player has to run 46.8 - 39.5953 m = 7.20475 m in 3.05427 s\nHis speed must therefore be 2.35891 m s^-1
834242
it hase more striking power that a straight ball..and also it is the lane conditions that they put out it is bone dry on the back ends anywhere from 15ft to 25 ft depending on they why that the PBA oils the lanes
325934
no, 34'' is to long and 33oz is to heavy for a standard game of softball that is not fast pitch. you should look into something more along the lines of 32'' 22oz because it's not fast pitch.
438060
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2