Question

Can someone explain 1B-2, e and f?

Answer 1

are you asking about the bouncing ball question?

For part e) I would modify the equation so that the max height is 1/2 of the original equation's max height.

From part b) you found the time to the max height is b/32
from part c, you found the max height = b^2/64
the result of part c suggests that "b" controls the max height.
if we create the equation
s= -16t^2 + b' t
and b' is b/sqrt(2) we will get a max height of
\[ \frac{b^2}{2} \cdot \frac{1}{64} \]
which is 1/2 of the original height

in other words, your new equation is
\[ s = -16 t^2 + \frac{b}{\sqrt{2} } t \]

Answer 2

the velocity will be the derivative of s, and you solve it the same way you did the first equation.

to find the time interval, solve for the t's that make s (height) = 0
you will get t=0 at the start and
\[ t = \frac{b}{16} \cdot \frac{1}{\sqrt{2}} \]

Answer 3

For part f:
Assume each bounce is 1/2 the height of the previous bounce. By the work of part e)
that means the new "b" is the previous b divided by sqrt(2)

we can make this equation:
\[ s= -16t^2 + b \cdot \left(\frac{1}{\sqrt{2}}\right)^k t \]
where k is the "bounce number", starting with k=0 (where we get the original equation)
and notice for k=1 , the second bounce, we get the correct b' = b/sqrt(2)
if we solve for the time interval, i.e. solve for t in
\[ -16t^2 + b \cdot \left(\frac{1}{\sqrt{2}}\right)^k t =0 \]
we get t=0 and
\[ t = \frac{b}{16} \cdot \left(\frac{1}{\sqrt{2}}\right)^k\]

and the length of the bounce is the difference between these times, i.e. the second expression.
now add up the times of each bounce:
\[\sum_{k=0}^{\infty} \frac{b}{16} \cdot \left(\frac{1}{\sqrt{2}}\right)^k \\
= \frac{b}{16}\sum_{k=0}^{\infty}\left(\frac{1}{\sqrt{2}}\right)^k \\
= \frac{b}{16}\sum_{k=0}^{\infty} r^k
\]
where r = 1/sqrt(2) , about 0.707
to sum this, see
https://en.wikipedia.org/wiki/Geometric_series#Formula