I don't understand how this problem is solved.

nombreuso

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So I have this problem:
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.
I have worked out a, b, c and d, which are b-32t, b/32 and b^2/64. At part d), it asks me to calculate where the second bounce is if the second bounce rises to half the height of the first bounce. In the answers sheet it says the second bounce is at b/16 + b/(16*sqrt(2)), but I don't understand this. Shouldn't it be at b/16, where the first bounce ends?
 
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At part d), it asks me to calculate where the second bounce is if the second bounce rises to half the height of the first bounce.

what is meant by "where the second bounce is" ?

maybe you should post part (d) as it is written in the problem statement
 
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what is meant by "where the second bounce is" ?
Actually, that isn't the question, it's just related to it. But in the answer sheet it says: "The second bounce is at b1/16 + b2/16. " and b2 = b1/sqrt(2).
 
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Question:
d) 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.)
Answer:
d) If the initial velocity on the first bounce was b1 = b, and the velocity of the second bounce is b2, then b2 2/64 = (1/2)b2 1/64. Therefore, b2 = b1/ √2. The second bounce is at b1/16 + b2/16.
 
first bounce has initial velocity [MATH]b_1[/MATH] and attains a maximum height of [MATH]\dfrac{b_1^2}{64}[/MATH] at time [MATH]t = \dfrac{b_1}{32}[/MATH]. Total time from the initial bounce to the second bounce is [MATH]2 \cdot \dfrac{b_1}{32} = \dfrac{b_1}{16}[/MATH].

second bounce has height equation [MATH]h = b_2 t -16t^2[/MATH]max height of the second bounce = [MATH]\dfrac{b_1^2}{128} = \dfrac{b_2^2}{64} \implies b_2= \dfrac{b_1}{\sqrt{2}}[/MATH]
 
first bounce has initial velocity [MATH]b_1[/MATH] and attains a maximum height of [MATH]\dfrac{b_1^2}{64}[/MATH] at time [MATH]t = \dfrac{b_1}{32}[/MATH]. Total time from the initial bounce to the second bounce is [MATH]2 \cdot \dfrac{b_1}{32} = \dfrac{b_1}{16}[/MATH].

second bounce has height equation [MATH]h = b_2 t -16t^2[/MATH]max height of the second bounce = [MATH]\dfrac{b_1^2}{128} = \dfrac{b_2^2}{64} \implies b_2= \dfrac{b_1}{\sqrt{2}}[/MATH]
Thanks, but I still don't understand what "The second bounce is at b1/16 + b2/16." means
 
[MATH]t = 0[/MATH] is when the ball initially moves upward at velocity [MATH]b_1[/MATH].
Hence, the first bounce after [MATH]t = 0[/MATH] is at [MATH]t = \dfrac{b_1}{16}[/MATH]
The ball's position equation after the first bounce is [MATH]s = b_2t - 16t^2 = t(b_2 - 16t) \implies b_2-16t=0 \implies t = \dfrac{b_2}{16}[/MATH]So, the second bounce occurs at time [MATH]\dfrac{b_2}{16}[/MATH] after the first, or [MATH]\dfrac{b_1}{16} + \dfrac{b_2}{16}[/MATH] units of time after [MATH]t=0[/MATH]
 
[MATH]t = 0[/MATH] is when the ball initially moves upward at velocity [MATH]b_1[/MATH].
Hence, the first bounce after [MATH]t = 0[/MATH] is at [MATH]t = \dfrac{b_1}{16}[/MATH]
The ball's position equation after the first bounce is [MATH]s = b_2t - 16t^2 = t(b_2 - 16t) \implies b_2-16t=0 \implies t = \dfrac{b_2}{16}[/MATH]So, the second bounce occurs at time [MATH]\dfrac{b_2}{16}[/MATH] after the first, or [MATH]\dfrac{b_1}{16} + \dfrac{b_2}{16}[/MATH] units of time after [MATH]t=0[/MATH]
So the second bounce is the third time the function equals cero?
 
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