Question

If the ball continues to bounce, how long does it take before it stops? i am not getting how it is
1/(1-(root2)) or 3.4 ?? help please

Answer 1

Can you post the full question?

Answer 2

1B-2 A tennis ball bounces so that its initial speed straight upwards is b feet per
second. Its height s in feet at time t seconds is given by s = bt − 16t
2
a) Find the velocity v = ds/dt at time t.
b) Find the time at which the height of the ball is at its maximum height.
c) Find the maximum height.
d) Make a graph of v and directly below it a graph of s as a function of time.
Be sure to mark the maximum of s and the beginning and end of the bounce.
e) Suppose that when the ball bounces a second time it rises to half the
height of the first bounce. Make a graph of s and of v of both bounces, labelling
the important points. (You will have to decide how long the second bounce lasts
and the initial velocity at the start of the bounce.)
f) If the ball continues to bounce, how long does it take before it stops?

how to solve the f (last part of the question) ?

Answer 3

From part a) we have the initial velocity = b (call it b0)
From part b) it takes t= b/32 to reach the max height
part c) It reaches a max height of b^2 / 64

If the ball reaches just ½ the height on the next bounce, then it left with a slower velocity (some of the speed is lost to friction)

Call the speed of the 2nd bounce b1.
the ball rises to height s = b1 t - 16 t^2
it has velocity v= b1 - 32 t
and reaches max height at time t= b1/32
the max height will be s= b1* b1/32 - 16 * (b1)^2/(32*32) = (b1)^2/64
we know this equals ½ the height of "bounce zero" , so we know
\[ \frac{b_1^2}{64}= \frac{1}{2} \frac{b_0^2}{64} \\
b_1= \frac{b_0}{\sqrt{2} }
\]
The time up to the max ht will be t = b1/32 = b0/(sqr(2)*32)
and double that for the time up and back down (a full bounce): b0/(sqr(2)*16)
\[ \frac{b_0}{16\sqrt{2}}\]
in general, the nth bounce will take time
\[ \frac{b_0}{16} \left( \frac{1}{\sqrt{2}}\right)^n\]

Answer 4

the total time will be the sum of the times of the bounces
\[\frac{ b }{ 16 }\sum_{n=0}^{\infty}\left( \frac{ 1 }{ \sqrt{2} } \right)^n \]

the summation is
\[ \frac{ 1}{1-r} \]
where r= 1/sqr(2)
see https://en.wikipedia.org/wiki/Geometric_series#Formula

we get
\[ \frac{ b }{ 16 } \frac{1}{1-\frac{1}{\sqrt{2}}} \\
= \frac{ b }{ 16 } \frac{\sqrt{2}}{\sqrt{2}-1} \\
= \frac{ b }{ 16 } \frac{\sqrt{2}\left(\sqrt{2}+1 \right)}{\left(\sqrt{2}-1 \right)\left(\sqrt{2}+1 \right)} \\
= \frac{ b }{ 16 }\left(2+\sqrt{2} \right)
\]

Answer 5

I think the given solution might be wrong.

given the position equation:
s = bt -16t^2
it is only necessary to solve it for s=0.
So we discover t=b/16.

On the second bounce b is now 1/2 what it was. So using the same equation substituting b/2 for b, we find t=b/32. and so forth. Assuming each sucessive bounce reduces the speed of the ball by 1/2

b is speed (velocities going up and down). There is no excuse for squaring it or taking the root.. The sum of the Zeno's paradox series is 1. So the time it keeps bouncing after it comes down the first time is 1 x b/16.

Answer 6

Well given the equation, s = bt - 16t^2
The derivative of ds/dt = b - 32t. This one is equivalent to its velocity

So the answer in a) is b(initial) - 32t.
In b however is when the velocity is 0 because the velocity at maximum height is 0.
So from the equation b(final) = b(initial) - 32t
We square b(final) to 0 so 0 = b(initial) - 32t
t = b(initial)/32
That's just it because you haven't stated the initial velocity.
For c) The maximum height is the height at time in answer b
So with the equation s = bt - 16t^2
s = (b(initial)^2)/32 - 16((b(initial))^2)^2
s = (b(initial)^2)/32 - 16(b(initial))^4
For d) You can do on your own I know it :) Also, you can also download this http://www.geogebra.org/cms/en/ . It helps you graph functions.
For e) Yeah you can also do it. Just set the 2nd maximum height as half of the first then conpute for the time using the previous equations.
For f) This is more of a geometric progression where the multiplier is 1/2. Basically this is just what Pi has stated in the previous answers :D