A ball is dropped 10 ft. from the ground. Then, it is left to bounce back and forth until it stops bouncing...
140637
I think the answer should be\n\n2 * 10/(1-4/5) - 10  = 90 ft\n\nas the new height would be 4/5 of the immediate previous height, so R = 4/5.\n\nand the sequence would be 10, 8, 32/5 ....
445917
it is 10 feet and it has been that way since before Wilt Chamberlain ever played which is about 65 or more years
873551
1st contact with floor\nS1 = 8\n\n2nd contact with floor\nS2 = 8 + 2*3/4*8\n\n3rd contact with floor\nS3 = 8 + 2*8*3/4 + 2*8*(3/4)²\n----------------------- ----------------------- ----------------------- \nnth contact with floor\nSn = 8 + 2*8*3/4 + 2*8*(3/4)² + ..... + 2*8*(3/4)^(n - 1)\n= 2*8  + 2*8*3/4 + 2*8*(3/4)² + ..... + 2*8*(3/4)^(n - 1) - 8\n= 2*8(1 - (3/4)^n)/(1 - 3/4) - 8\n= 64(1 - (3/4)^n) - 8\n\neg After 10th bounce\n\nS10 = 64(1 - (3/4)^10) - 8\n≈ 52.4m\n\nThe limiting distance is 64(1 - 0) - 8 ie 56 m
893929
Fg = mg = (0.20 kg)(9.80662 m/s^2) = 1.961324 N\nFc = (mv^2)/r = mrω^2\nFc = (0.20 kg)(0.60 m)((2π rad/s)^2\nFc = 4.73741 N\nFt = √( 4.73741^2 + 1.961324) @ arctan(1.961324/4.73741)\nFt ≈ 5.127 N @ 22.5°\n\n[The vertical component of Ft must offset the weight of the ball.]
78760
a basketball will bounce the highest of these objects because it is mad of a harder material whose material will retain its shape more readily than the others. also it has higher air pressure inside.
495581
a round ball I guess LoL
1069779
Draw a free-body diagram of the ball and examine it's accelerations - if you assume wind and air resistance are negligible and that the ball is accelerated straight up, then you're left with its initial acceleration (68.57 m/s^2) and gravity (9.8 m/s^2 if I remember correctly) \n\nThe ball will decelerate by 9.8 m/s every second, you can divide to find the amount of time the ball takes to reach its apogee (point of zero acceleration), then integrate the acceleration curve to find the velocity curve, which can be integrated again to find the distance traveled.  I'm not sure it can be done without the calculus.
608626
Due to law of conservation of energy. The kinetic energy it gains during falling is restored in the rubber at the time of collision while in other balls it is lost in sound & heat. So the crazy ball bounce much more than other balls.
517842
Keep your eye on the ball.
486244
The ball and the basket.
957111
The angle is given. Horizontally means with an angle of 0 degrees elevation.\n\nYou can find how long the ball takes to travel 60 ft horizontally (60/147.84 seconds), so, knowing the acceleration due to gravity (which you will have been given), you can calculate how far it must fall in that time.\n\nt = 60/147.84 = 0.405844156\n\ndistance d covered by falling body in t seconds = ½gt²\n\ng = 32.174 ft/sec²\n\nd = 0.5 * 32.174 * 0.405844156 * 0.405844156 = 2.649681386 ft
894241
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.
929319
32 feet per second per second.  The acceleration is constant while the ball is in the air.  It is determined by the force of gravity.  The ball's speed will vary, but acceleration is constant.
580059
If you take a ball and hold it out at arms length, it has potential energy. This energy is being held back by your hand and once you release the ball, gravity turns that potential energy into kinetic energy (movement) and makes it hit the floor.
470687
At the top of the halfpipe, the ball is just sitting there with potential energy stored up (PE = mgh, where m = mass, g = acceleartion due to gravity, and h = height).  As the ball rolls downward, its PE lessens and its kinetic energy increases (KE = 1/2(mv^2), where v = velocity of the ball).  At the bottom of the halfpipe, all PE is gone and KE is at maximum.  As the ball goes up the other side, the KE decreases and the PE increases.  In a perfectly frictionless halfpipe, the ball would return to the same height as it was before it started rolling, and the KE would go to 0 and the PE would be at its maximum.\n\nIf you just dropped the ball (no halfpipe - just a freefall), the KE just before impact would be the same as the PE before dropping.  The two are inversely proportional - as one goes down, the other goes up.\n\nNow if you knew the exact shape of the halfpipe, you could start plugging in numbers to find the PE and KE of the ball at various points.
895329
When you hold a ball up in the air, it develops potential energy because gravity wants to pull it down.  As the ball is dropping, that potential energy became kinetic energy, but loses some energy because of friction with the air.  At impact more energy is lost to the ground via thermodynamics (heat loss) and deflection (strain energy transfer).  The ball bounces back up with the remaining energy.  Something like that, better to ask a physics professor in high school or college :)
705043
As the ball hits the ground, its momentum causes its elastic body to compress itself against the ground until enough force had been applied through enough distance to cancel the momentum of the ball and bring it to a stop.  At the instant the ball comes to a stop, it has energy stored in it due to the compression of its body against the ground.  This compressive energy is then released in pressing against the ground, causing the ball to accelerate upward.