Question

Need help understanding the answer for Problem Set 1B-2 part F

Question is:

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)

F) If the ball continues to bounce, how long does it take before it stops?

Answer 1

Solution is: \[\frac{ b/16 }{ 1-\sqrt{2} }\] due to the geometric series.

I understand that it would become a geometric series, i just don't understand the math behind it. For instance, why is \[b _{3} =\frac{ b }{ 16(\sqrt{2})^{2} }\]

Answer 2

The formula you've given is not the formula for b_3, but rather the formula for the landing time for the third bounce. The formula for b_n is\[b _{n}=\frac{ 1 }{ \sqrt{2} }b _{n-1}=b \left( \frac{ 1 }{ \sqrt{2} } \right)^{n-1}\]Therefore we get\[b _{3}=b \left( \frac{ 1 }{ \sqrt{2} } \right)^{2}\]For each bounce, the landing time is (b_n)/16, and the formula at the end of your post comes from rearranging (b_3)/16.

By the way, the formula you gave for the solution contains a typo. You would need to replace sqrt(2) with 1/sqrt(2).

Answer 3

Right. I'm trying to understand how you got the formula for b_n.

My question is:

1) How does \[b _{n} = \frac{ 1 }{ \sqrt{2} }b _{n-1}= b (\frac{ 1 }{ \sqrt{2} })^{n-1}\]
I will definitely try to work out how and why the formula for b_n is that but just incase I can't figure it out, I would appreciate if you could explain your thought process.

The thing that makes sense to me is;

If the second bounce occurs at \[\frac{ b _{2} }{ 16 }\] and\[b _{2} = \frac{ b _{1} }{ \sqrt{2} }\] then it makes sense that the second bounce in terms of the b_1 is equal to \[\frac{ b }{ 16\sqrt{2} }\]

Thanks for the help thus far.

Answer 4

By the way, do you think I should brush up on some concept before I tackle this?

Answer 5

I don’t think you need to brush up anything except maybe practice tennis a little.

We start out with the information that the initial velocity of the bounce is b. When we start looking at a series of bounces, we designate this quantity as b_1, so we have\[b _{1}=b*1\]We’re told that the second bounce is half as high as the first one. The height of the first bounce is (b_1)^2/64, and we can use this information to find b_2, the initial velocity of the second bounce:\[\frac{ (b _{2})^{2} }{ 64 }=\frac{ 1 }{ 2 }\frac{ (b _{1})^{2} }{ 64 }\]\[(b_2)^2=\frac{ 1 }{ 2 }(b_1)^2\]\[b_2=\sqrt{\frac{ 1 }{ 2 }(b_1)^2}=\frac{ 1 }{ \sqrt{2} }b_1=b*\frac{ 1 }{ \sqrt{2} }\]The third bounce is half as high as the second one. The height of the second one is (b_2)^2/64. We can go through exactly the same steps above to discover\[b_3=\frac{ 1 }{ \sqrt{2} }b_2=\frac{ 1 }{ \sqrt{2} }\left( \frac{ 1 }{ \sqrt{2} }b_1 \right)=b*\left( \frac{ 1 }{ \sqrt{2} } \right)^2\]I hope this makes the connection for you but if not let us know and we'll try again.

Answer 6

Okay, I over-thought this one and in the wrong direction lol.

My algebraic equation was just completely off.

Thanks again, Creeksider!