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Vera_Pavlovna [14]
3 years ago
14

Suppose you drop a tennis ball from a height of 6 feet. After the ball hits the floor, it rebounds to 80% of

Physics
2 answers:
statuscvo [17]3 years ago
7 0

Answer: The height after the third bounce is 3.1 meters

Explanation: We know that after every bounce the ball rebounds to 80% (or 0.8 in decimal form) of its previous hight. So if the first height was 6m, after the first bounce, the height will be:

H1 = 6m*0.8 = 4.8m

after the second bounce we have:

H2 = 4.8m*0.8 = 3.84m

and in the third bounce, the height is:

H3 = 3.84m*0.8 = 3.072m

Now we need to round it to the nearest thenth, this is the first number after the decimal point.

We can see that the second number after the decimal point is a 7, so we should round up, and get H3 = 3.1m

Another way to model this situation is with the equation:

H(b) = 6m*0.8^b

Where b is the number of bounces that the ball did.

in this situation, the third bounce is when b= 3, and we have that:

H(3) = 6m*0,8^3 = 3.072m, which is the same result as before.

sasho [114]3 years ago
3 0
The formula to solve this is (.80)^3 X 6 and the answer would be 3.1 feet. That is how high the ball will rebound after its third bounce. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
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Answer:

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We know that,

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Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

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Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

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