Bouncing ball and momentum

taveuni

Hi:

I am working through this problem regarding a bouncing ball. I have gotten most of the way through, but can't figure out the time interval between bounces. Can someone give me some guidance? What am I not seeing?

A 32 g steel ball bounces elastically on a steel plate, always returning to the same maximum height h = 20.2 m.
a) With what speed does the ball leave the plate?
vup = 19.9m/s
b) What is the magnitude of the total change in momentum of the ball with each bounce?
|Dpball| = 1.27
c) What is the magnitude of the total momentum the ball transfers to the plate at each bounce?
|Dpplate| = 1.27
d) What is the time interval between bounces?
Dt = ???

I tried to simply t=d/v=20.2/19.9 and multiplying that by 2-- wrong. I can't think of any other way to solve this. And the next question builds off of it:

e) Averaged over a long time interval, what is the magnitude of the rate at which momentum is transferred to the plate?
|Dp/Dt| =
f) What is the magnitude of the average force exerted by the ball on the plate?
Favg =
Thank you...

Office_Shredder

Time interval between bounces...

knowing the laws of kinematics, can you calculate how long it takes for an object to move a set distance, given the acceleration?

taveuni

I know
dx=vo*t+(a*t^2)/2
But that doesn't give me the right answer
I got 0.84s ---
??

taveuni

Is there something I am completely missing here?
dx=vo*t+(a*t^2)/2
20.2 = 19.9t + 9.81/2*t^2
solving for t = 0.84 or -4.9
2*t = 1.68s. Wrong.

Please, does anyone have any suggestions?
Thank you so much.

goodboy

20.2=9.81/2*t^2,=>t=2.03
or 20.2 = 19.90792t - 9.81/2*t^2

PPonte

[tex]v = v_0 + at[/tex]

[tex]v - v_0 = at[/tex]

[tex]t = \frac{v - v_0}{a}[/tex]

v - velocity of the ball when reaches the maximum height
v₀ = vup = 19.9m/s
a = g = - 9.80 m/s²
t - That's the time the ball will take to reach the maximum height.

But the time interval between two bounces let's call t_b is equal to 2t.
Then:

[tex]t_b = 2(\frac{v - v_0}{a})[/tex]