Free Fall

What do you mean by "the ball is in flight for 5 sec?" Is that the total time bouncing? Under normal conditions, the ball will reach the ground in well under 5 seconds?

If the ball always rebounds to half the previous height, then theoretically, it never comes to rest. It will, however, travel a finite distance.

The ball is accelerating after it is dropped. At what point do you want the speed?

ok
I don't know what the question is asking for; it was written by my teacher.

i can help you.im not a american,so i might not say the words clear enough.first,the air is acting on it to slow it down,Q=k*l(l represent the length the ball travels in the air) second,the total energy the ball has is E=mgh(h the height),and the time it comes to a rest is the moment all the E is translated into heat.E=Q does it make any sense

An object dropped from 190 cm with negligible air friction will fall for the following amount of time.

s = (1/2)gt^2

(2s/g) = t^2

t = SQRT(2s/g)

g = 9.8 m/s^2 * 100 cm/m = 980 cm/s^2

t = SQRT(2(190 cm)/980 cm/s^2) = SQRT(380/980 s^2)
= SQRT(0.388 s^2)
= 0.6227 sec.

This, not 5 seconds, is the amount of time that it will take the ball to reach the ground. I am not sure what this question means.

When the ball strikes the ground, its speed will be 0.6227 sec(980 cm/s^2) = 610 cm/s.

hey
Thank you very much.

Sorry, I caught an arithmetic error. 9.8 * 100 = 980 not 9800. Hope the correction isn't too late.

hey
No it isn't too late.

Thanks