how does the temperature of a squash ball affect the height of the bounce?
402998
This is an interesting question.  I think tyba has the right idea.  I would say that higher temperatures lead to a more flexible ball, which in turn will lead to a more elastic collision with the ground.\n\nJust in case you don't know:\nA perfectly elastic collision is one in which the object has exactly equal and opposite momentum after the collision (no loss of kinetic energy).  A perfectly inelastic collision is one in which all kinetic energy is converted to another type of energy.  No real collision is either one of these, but rather somewhere on the spectrum between them.\n\nIn the case of the frozen ball, I would say that most of the energy is converted to thermal energy (heat).
833124
I just got into playing squash with a buddy. It's a great sport. Very fun and  it's good cardio.
800626
If you are asking: \n"What is the relationship between volume of a ball and it's height (diameter)?", the answer is V=(4/3)r^3=(D^3)/6 where D is the Diameter.\n\n"How the size of a ball affects the height a ball will bounce back when dropped?"  There are several variables, each of which would need to be addressed.  First, is this a real world activity, or a theoretical "perfect" situation?  Given that answer, are the balls striking the same surface, are they made of the same materiel, are the of the same or different masses, are they in the same gas/gravitational field/barometric pressure/temperature/(EM flux), what is range of sizes you wish to compare (relative vs absolute)?\n\nThe real answer here is that you need to clarify the question, as there are about 16 different interpretations of the question not listed, each with their own variable set to consider.  If this is a homework question, just read the lesson.  IF it is a real question of your own, please rephrase it more clearly.
64850
A correction to the first answer. The formula terminal V = SQRT(2W/(Cd*R*A)) is correct, as it expresses the fact that W is balanced by the drag force V^2*Cd*R*A/2. However, for two same-sized smooth balls of different materials, only W changes; Cd stays the same. The air density R depends only on altitude.\nSince air density increases with decreasing altitude, the terminal speed decreases during a fall. If dropped from high enough, the actual fall speed, which initially = 0, increases until the (decreasing) terminal speed is reached, and then the sign of the net force (gravitational + drag) changes so the falling ball slows. For the remainder of the fall, the terminal and fall speeds both decrease, but the fall speed is slightly faster. This speed difference provides the net upward force to slow 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.
14589
My guess is 2 things... First of all, the lower the air pressure, the less resistance that the ball will encounter along its path.  Secondly, if the ball is hollow (e.g. a tennis ball or basket ball) then the pressure difference between the inside and outside of the ball will reflect how easily it will bounce.
601942
It needs to be adjust by the height of each player. It should be just high enough that the ball is at belt level on each player.
142406
Mass is not affected by temperature, but volume can be slightly changed because if an object heats up, the particles begin to move faster and so it expands, or vice versa if the object gets cooler.
249004
Squash is played between two individuals (singles) or two teams of two (doubles) on a Squash Court. \nOne player serves to start a rally which proceeds until one player hits the ball out or down or fails to hit the ball before it has bounced twice. \nSquash can be played using  a couple of scoring options, traditional (International) scoring and point a rally scoring (PARS). \nSquash Players having a social game can modify the rules to their own requirements. \n\nIf your fitness is poor or you haven’t played Squash for some time, don’t overdo it. \nRemember, Squash is a fast game.  If you have a rally which requires a lot of short sharp movement, it may take your body a moment or two before it realises that a lot of oxygen is needed for recovery. \nThis can result in a sudden increase in heart rate. \nThis is not a good thing for those who lack fitness\n\nCheck the link for much more info!\nGood luck!
379193
it has nothing to do with the air in the atmosphere its the air that fills the balloon up. when its hot the ballon goes up as it cools it goes down
573834
Its potential energy was 78.4 m joules before falling, where m is the mass of the ball.\n\nAfter collision when it reaches the height 5m its potential energy is 49 m joule only.\n\nIt has lost 8m joule of energy due to inelastic collision.
1164981
I think you have left off part of the question. The speed of a ball is a vector. When you hit the ball, the vector points upward and forward. Gravity pulls the vector downward. Small forces (air friction, etc.) retard the total speed, but assume those can be ignored.\n\nAt some point, the pull of gravity will pull the vector to point down instead of up. That's the maximum height. The horizontal component of the speed vector should not change (given our assumptions). It still goes forward at the initial velocity. However, The vertical component is instantaenously at zero. In a moment, the ball will start to accelerate vertically and the vector will continue to pull down below the horizontal line becuase of the force of gravity.\n\nIf this were not true, the ball would reach maximum height and then just drop from the sky (0 overall velocity) which you know it does not do.\n\nSo of your choices, probably equal to initial or ever so slightly les than initial would be the best.\n\nThe same thing happens if you drop, say, a ball from 10 feet and fire a bullet from a height of 10 feet. The both hit the ground at the same time.The fact that the bullet is speeding horizontally doesn't affect gravity pulling on it vertically. (yeah small effects like curvature of the Earth, wind resistance, etc. being ignored).
255789
The only way you are going to know for sure is to try it.\n\nIt will be interesting to make some predictions about what will happen. Will it be smaller, less bouncey, more squashy, etc.? And Why.
710073
They had that on an episode of Mythbusters.  The ball travelled less further because of the decrease in mass and no change in volume.  It was more susceptible to air resistance.
879264
'Large' and 'small' probably refer to the masses of the balls, rather than the sizes. Assume one ball (with mass m1) is initially directly above the other (mass m2), and both are dropped at the same time from rest (with a slight vertical space between the two).\n\nThe lower ball hits first, bounces up and collides with the upper ball near the ground. The two  have about the same speed v = sqrt(2gH) going into this collision because both balls have fallen from approximately the same height. If the two masses were equal, then both would simply reverse direction, the top ball going upward at speed v and the bottom ball going down (for the second time) and immediately up at the same speed v --> both balls would continue up at the same speeds to the original height H.\n\nNow suppose the masses are slightly different with m2 > m1. Then the initial and final velocities (taking the upward direction as positive) for the collision are respectively:\n\nv1 = -v\nv2 = v\n\nand:\n\nv1 = v + x1\nv2 = -v + x2 \n\nwhere x1,x2 << v. Energy and momentum conservation require that:\n\n(m2-m1)v = m1(v + x1) + m2(-v + x2)\n(m1+m2)v^2 = m1(v + x1)^2 + m2(-v + x2)^2 \n= m1(v^2 + 2vx1) + m2(v^2 - 2vx2) (Approximately)\n-->\n2(m2-m1)v = m1x1 + m2x2\n0 = 2m1vx1 - 2m2vx2\n-->\nx1 = (m2/m1 - 1)v\nx2 = (1 - m1/m2)v\n\nSo the top (lighter) ball ends up with a higher speed (m2/m1)v and the lower (heavier) ball ends up with the lower speed (m1/m2)v. This is reflected by the one bouncing higher than H and the other lower than H.\n \nThere can't be more than one collision since the top ball comes out of the first (and only) collision with a higher speed than the bottom ball so the latter can never catch up to it. If the mass ratio were reversed (heavier ball on top) the situation would be much more complicated since there could be multiple collisions.
526447
I actually had my class conduct an experiment on this very thing.  The problem was: does the percentage of energy lost from a ball bouncing increase, decrease, or stay the same as the height increases.  This is akin to your question, because the height at which it bounces back to depends on how much energy is left after the bounce.  \n\nWe found that the percantage of energy lost stays a constant, which means the the answer to your question is yes.\n\nFor example:\nIf you start with a height of 10 meters and drop a 1.0 kg ball, it will have 98 Joules of potential energy (PE = mgh).  If it loses 10% of its energy on the bounce, it will have 88.2 Joules of kinetic energy, which will translate to a height of 9.0 meters.  \n\nIf you double the original height, you will have 196 Joules of energy.  After losing 10% from the bounce, you are left with 176.4 Joules, which translates to a height of 18 meters after the bounce.
664646
just make it really tall, cause the ball goes pretty high but then again it cant do any real damage to it.