What is the equation to find the height of a bouncing ball under earth's gravity (9.8?) if given the time (t) from the start of the drop (10ft) if the ball is either a tennis ball or a ball that reaches 1/2 of the previous max height? (Ignoring air resistance & spin) In addition, the angle of drop needs to equal 90^{o} from a perfectly level and flat ground. Ohh and I would like to know the name so i can look it up on the web... Thank you for any help....
Just use conservation of energy. mgh=1/2 m v^2 solve for the velocity just as the ball hits the ground. Then use conservation of momentum and the coefficient of restitution of the ball to determine its velocity after the bounce. Then use energy again to solve for the new height.
Welcome to PF! Hi TECH GEEK! Welcome to PF! The equation you need (between bounces) is one of the standard constant acceleration equations, s = ut + at^{2}/2. s is distance, u is the initial speed (in this case zero), t is time, and a is acceleration (in this case, 32 ft/s^{2}). And if the height is 1/2 the first time, it will be 1/4 the second time, 1/8 the third time and so on. (and I think the best search-word would be "projectile")
What if i want to figure for a tennis ball? (Mass = 58 grams, max height of 2^{nd} bounce = ???, height of drop = 10ft) Rest of conditions are same as before... By the way, I need to make a graph of the results.
To find the time, t, to drop 10 ft from rest, the mass is irrelevant, and so is the height of the subsequent bounce. The equation is 10 = 32t^{2}/2.
What I meant by the three colored parts of this quote, is that... #1. "Height" = The height from the ground to the point of reference on the ball... ((This is basically the position at which the ball will be at the singular point of time given)) -Note: See "#2" for "Time given"- #2. "If given the time (t)" = Time after the start of the drop... (I.e. If the point in time the ball is dropped = 0, and "t", or time = +24 sec's, then logically "t" is the equivalent of 24 seconds after the drop) #3. "(10ft)" = Change this to 100cm...
Ok so let H be the initial height. The potential energy in the system at the beginning is mgH. So you can solve for the velocity of the ball just as it hits the ground by using conservation of energy. All of the potential energy becomes kinetic energy. mgH=1/2 m V^2 so V=sqrt(2gH) Now to determine how high the ball will bounce you need one of two things. You either need to assume a perfectly elastic collision in which case no kinetic energy is lost and the ball will begin to travel upwards at the same velocity we just solved for, and therefore will reach the initial height. This is however unrealistic so we need to know the coefficient of restitution between the ball and the floor, lets call this "e" and it is a measure of how bouncy the ball is. The coefficient of restitution is between 0 and 1 with e=0 meaning the ball will not bounce and e=1 meaning you have a perfectly elastic collision. http://en.wikipedia.org/wiki/Coefficient_of_restitution So using the coefficient of restitution we can then solve for the velocity upwards. Lets say V1 is the downwards velocity we solved for above, and V2 is the new upwards velocity. V2=-eV1 (negative because it is in the other direction) Then we can use conservation of energy again to determine the new height. 1/2 m (V2)^2 = m g H2
I did some tests on a new tennis ball... It's c.o.r. was about 0.75... Any more help is gladly apprecreated!!!!!!
As tiny-tim said, the formula for the height of the ball is [tex] h = h_0 - \frac12 gt^2 [/tex] This is the formula for a parabola. But this equation only works during free-fall. So each bounce is governed by this equation. But when the bounce occurs it loses some of its energy and velocity. What is the velocity of the ball right before it hits the ground? Use energy like RandomGuy88 said: [tex] \frac12 mv^2 = mgh \implies v_f = \sqrt{2gh_0} [/tex] What is the velocity after the bounce? Use the COR: [tex] v_i = (COR)(v_f) [/tex] So this is the initial velocity of the second bounce. How high does it get? Again use the conservation of energy, but the other way around: [tex] \frac12 mv^2 = mgh \implies h_0 = v_i^2 / 2g [/tex] This is how high up the second bounce gets. What is the formula for this flight? Same as before: [tex] h = h_0 - \frac12 gt^2 [/tex] Then it starts to fall again...and the problem repeats. Again and again.
The ball is key, the coefficient of restitution is the kinetic energy the ball will exert given the height and weight of the ball and what the ball is made of. The equation is useless because it will not be valid without the joules exerted from the impact. Weight is necessary because this will be the main fact in calculating joules from velocity. What its made of is important to calculate the exchange of joules and what joules would be conserved. Tiny tim shows you the equation for terminal speed on impact, but the formula to calculate the height of the bounce needs more information. If you want to learn more google kinetic energy or coefficient of restitution.
The coefficient of restitution--as several posters have mentioned already--is the ratio of the ball's speed after colliding with the floor to its speed before colliding. Look it up.
Coefficient of restitution COR or the bounciness of an object is the objects potential to transfer joules. Don’t bother me with this general observation. I could say that angular momentum would be the ratio of height lost over time after impact but I would rather call it a parabola. Saying one ratio or variable is more important than the other when calculating a reaction is called nit picking. Kinetic energy is not just calculated with coefficient of restitution. Saying restitution potential would be the ratio gains-base recovery. But the coefficient of restitution is the objects potential to transfer energy, kinetic energy that is. Sorry to nit pick. TM
I could say you need to calculate the coefficient of friction, it’s going to help you just as much as coefficient of restitution. Or what about static friction in the ground being sand, concrete ,wood. This would affect the coefficient of restitution. So would that be the ratio of potential restitution and kinetic absorption due to static friction of the two bodies. Again sorry for nit picking. TM