an athletic with 1/70 m tall, wants to throw a ball to the basket from penalty spot .indicate under wich angle & wich speed this ball can go to the basket .(the distance between spot penalty till basket is 5/80)(the distance between ground till basket is 3/05)
218312
There are infinit combinations of speed and angle to do it. You shold set a speed that will be constant and calculate the angle taking care of the gravity acceleration, depreciating air resistivity, the ball will make a parabolic way to the ring...
92103
The height of the net shall be 2.43 m (7’ 11 5/8”) for men and 2.24 m (7’ 4 1/8”) for women.\n\nThe net height is measured from the center of the playing court with a measuring device. \n\nThe two ends of the net (over the sidelines) must both be at the same height from the playing surface and must not exceed the official height by more than 2 cm (3/4").
223160
put the ball in the hole
309853
It's all about conservation of energy.   The contestant did some work on the ball.   Remember the definition of work?\n\nWork = Force * Distance\n\nSo, the contestant did a work:  \n              W = 810N * 2.0m = 1620Joule\n\nOk, since energy is conserved, the ball will have a kinetic energy of 1620Joule at the begining of the throw, but it will eventually lose it as it goes up, transformed into Potential energy.\n\nWhen the ball reaches the highest point its speed is equal to zero, and hence the kinetic energy is zero too.  Conservation of energy requires that all 1620Joules transformed into potential energy.  So, do you remember the formula for potential energy?\n\nP. energy =  mass * gravity * height ; \n\nwith gravity = 9.82m/sec^2\n\nSince the potential energy is equal to 1620Joules:\n\n1620Joules = (7.0Kg)*(9.82m/sec^2)*height\n\nSo:\n\nheight = (1620Joules)/(7.0Kg * 9.8m/sec^2) = 23.567m\n\nHope I helped
406021
player A.  the higher vertical makes me assume he's more athletically-gifted since it takes both upper & lower body strength to jump higher.  player B may be foul-prone since he'll swipe at balls knowing he has the longer reach vs. player A, and he'll be more tired at the end of games from having the longer, heavier arms.  the more proportional the body of a player, the better his balance...which may translate to better shooting & dribbling.  :)
23962
This is a standard:\n\nassume by absurd that there are two different limits p0 and p0': call d the distance between them:\nd = |p0 - p0'|\nNow try to figure out: for any E there must be a N such that all points pn for n>N drop into the ball of radius E and center p0\nBut for THE SAME E there must CONTEMPORARILY be a N' such that for any n>N' all points pn drop into the ball of radius E and center p0'\n\nKeep the greater between N and N'; keep E=d/3 such that the two balls don't intersect in any point, and get to a conclusion\n\nGood work and good luck!
559768
have fun with that. i live in america as a 6' sg/pg and shoot over 50% from 3. the most ive done is talk to a d2 coach or two. but fill out the ncaa clearinghouse info and just look on college websites and email coaches. that and make a video of yourself playing real games and send that to them.
590101
The ref was correct, in my opinion.\n\nHere are some of the pertentent free throw rules for the NBA (I know, not college or high school, but I believe that they are similar):\n\na. After the ball is placed at the disposal of a free throw shooter, his attempt shall be within 10 seconds in such a way that the ball enters the basket or touches the ring before it is touched by a player. The shooter shall be within that part of the free throw circle behind the free throw line. \n\nc. A player who occupies a free throw lane space shall not touch the floor on or across the free throw lane line, nor shall any player 'back out' more than 3' from the free throw lane line. A player who does not occupy a free throw lane space must remain behind the three-point line. This restriction applies until the ball leaves the free thrower's hands. \n\nThese two rules might be slightly different for college or high school ball, but I am not sure.\n\nNow, here is the ruling on penalties levied:\n\n(6) If there is a free throw violation by each team, on a free throw which is to remain in play, the ball becomes dead, no point can be scored and play shall be resumed by a jump ball between any two opponents at the center circle. \n\nThese are NBA rules, but I believe that, though the mechanics might be a little different (jump ball, for example), the generalities are still the same for college and high school.\n\nThe shooter and the defender both violated rules, and there is no precendent for one rule or the other, and so the ball becomes dead, perhaps, in this case, being awarded to the other team?
535039
Well, let's start here.\n\nSTEP 1: Award 1 point for each free throw made. Free throws (also called foul shots) are attempted after a foul has been committed on a player who was in the act of shooting. \n\nSTEP 2: Add on 2 points for each successful basket made against the opposition. \n\nSTEP 3: Award 3 points for each successful basket made from beyond the 3-point line, which in the National Basketball Association (NBA) ranges from 22 feet to 23 feet 9 inches from the basket. A 3-point shot in the Women's NBA is 19 feet 9 inches from the basket. \n\nSTEP 4: Note that additional points are awarded in various other situations. Technical fouls, illegal defense fouls and flagrant fouls often result in free throws. \n\nSTEP 5: Understand that the team with the most total points at the end of regulation time wins the game. Barring overtime, NBA games last 48 minutes and college games last 40 minutes. Tips & Warnings \n\nAlso what level of basketball do you keep score for? This may help you in your journey in terms of keeping score?
371518
Let's do question 2 first since this will make it easier to see how to do question 1.\n\nLet H be home, A be first base and B is 2nd base.\n\nThe angle HAB = 90 degrees\n\nlet x ft. be the player's distance from first base and let y be his distance from home, then from the pythagorean theorem we have:\n\n(1)   y = sqrt(90^2 +x^2) ft = (90^2 +x^2)^0.5 ft\n\nWe are given dx/dt = 28ft per sec (a constant. Acceleration is ignored.)\n\nNow: (2)   dy/dt = (dy/dx)*(dx/dt)\n\nWe are given dx/dt and we can determine dy/dx by taking the derivative of equation (1), using the rule for exponents and the chain rule:\n\ndy/dx = 0.5(90^2+x^2)^ -0.5 * d/dx (90^2+x^2) \n\nNow: d/dx (90^2+x^2) = 2x, so\n\ndy/dx = x / (90^2 + x^2)^0.5\n\nWhen the player is 30 ft from 2nd, he is 60 ft from first. So x = 60ft.  We can now determine dy/dt from (2) using x = 60 ft\n\ndy/dt = (60 /(90^2+60^2)^0.5)*28 ft/sec\n\n= 15.53 ft/sec\n\n(note: this problem can also be solved using angular measure and simple trig)\n\nWhen doing the first problem, label 3rd base C, and then the angle HCB = 90 degrees, and then solve it in the same manner.
460460
i don't think the ball is bouncing in this question.\n\nassuming that the ball falls off from a very tall building, initial speed = 0 m/s\n\ndistance the ball falls in the first second = ut + 1/2 at^2 = 1/2 a t^2 ( since initial speed is 0 m/s)\n= 0.5*9.81*1*1 = 4.905 m\n------------------------------------------------------------------------------------\n\nafter the ball falls for 1 second, the speed of the ball is 9.81 m/s (since acceleration of free fall is 9.81 m/s^2\n\ntherefore, distance ball falls in the second second\n= 9.81*1 + 0.5*9.81*1*1 (using the same formula as above)\n= 14.715 m\n\nusing the same process you can find the distance which the ball falls through for the third second.
98833
For a back row spike I found 26.7-27.6m/s (mean).  Sorry, but that was all I could find!  I was going to do a study on it - but then I switched out of biomechanics into exercise physiology.  You could do it by calculating time elapsed between contacting the ball and the ball reaching the floor.  You would have to know the height at which the ball was contacted and how far it traveled horizontally from the point of contact.  They you find the hypotenuse using Pythagorean theorem and divide this number by the time to get an estimated velocity.
486427
answer is approxomately 42.476m/s\n\nonly one formula here \nS=ut + (at^2)/2 \ns- distance travelled = 40m\nu- initial valocity\na - acc due to gravity-9.8m/s^2\nt- time \nin the first case the initial velocity is 0m/s^2\nso\n40=(9.8*t^2)/2\nt=2.857 sec\nas both the balls have to touch the ground at same time  and the second ball being dropped( rather thrown would sound correct) after 2 secs, the later has only 0.857 sec left for traelling a distance of 40m. \nso \n40=u*0.857+(9.8*(0.857^2))/2\nsolving u=42.476m/s\nthe initial velocity for the second ball should be 42.476m/s and should be thrown vertically downwards.\nthat is pheta=0 in the formula given un the question.
233478
A: The ball will be at a height of 6ft when y=6\nSo -16t^2+32t=6 dividing by -2\n8t^2-16t+3 =0\nSolving for t\nt1 = (16+sqrt(256 - 4*8*3)/(2*8) = 1.790569415\nt2 = (16-sqrt(256 - 4*8*3)/(2*8) = 0.209430585\nB: Put y = 5\n\nSo -16t^2+32t=5\n   16t^2 - 32 t +5 = 0\n\nt1 = (32+sqrt(1024 - 4*16*5)/(2*16) = 1.829156198\nt2 = (32-sqrt(1024 - 4*16*3)/(2*16) = 0.170843802\nThe kicker distance from the goal \nEither 8.54219012 ft \n   Or 91.4578099 ft  ( the right answer )
232451
Yes, defensive three seconds. If you aren't within arms length of an offensive player in a defensive position (knees bent, protecting the post, etc.) you will be called for defensive three seconds. The penalty is one technical foul shot for the offensive team and the ball in a sideline out of bounds.
326253
When I play pick-up games with my friends, I like to work the post too.  I just usually do double-pumps, spins, and just kick it out when I can't get anything going.  Just try sweeping hooks and they should help.  A fadeaway turnaround is good when you don't want to get blocked.  It's a good move to have but I would say work on your hook shot because that thing comes in handy.  How tall are you and what level do you play (h.s., college, street, etc.)?
531933
D = d0 + d sum(%b^N) for N = 1 to 5 bounces; where d = 12 ft. the initial height and %b = .6 and d0 = (12 or 24) depending on if raising the ball to height d = 12 is included in the total distance.\n\nYou can show this by:\n\nN = 0 the starting point with d0 = (12 or 24 ft.), depending on if you count raising the ball to 12 ft. as well as dropping it to start the bouncing.\nN = 1 the first bounce; so (.6)^1 = 60%; so the distance it travels here (d1) is d (%b) up AND also d (%b) down to start the next bounce.  Thus, d1 = 2 d (%b^1)\nN = 2 the second bounce; so  (.6)*(.6) or 60% of the 60% first bounce; so the distance d2 = 2 d (%b)^2\nN = 3 the third bounce distance would be similar; so d3 = 2 d (%b^3) and so on up to N = 5.\n\nIn which case, D = d0 + d1 + d2 + d3...+ d5 = d0 + 2 *12(.6 + .6^2 + .6^3 + .6^4 + .6^5) = d0 + 2 * 12 sum(.6^N) for N = 1 to 5; If you count raising the ball to the initial height, then d0 = 24 = 2 d; otherwise d0 = 12 ft., the distance of the first drop.\n\nYou can do the arithmetic.