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4 years ago

A ball is dropped from a height of 32 feet. it bounces and rebounds 80% of the height from which it was falling

Physics

2 answers:

n200080 [17]4 years ago

7 0

Answer:

d = 32*( 1 - 0.8^n) / 0.2 = 160*( 1 - 0.8^n)

Explanation:

Given:

- The initial height of the ball h_o = 32 ft

- The successive decrease in height after every bounce = 0.8*h

Find:

- The expression relating the total distance traveled by the ball for nth number of bounce.

Solution:

- The distance traveled by the ball upto n = 1, is = 32 ft

- The distance traveled by the ball upto n = 2, is = 32 + 2*32*.8 = 83.2

- The distance traveled by the ball upto n = 3, is = 83.2 + 2*32*.8*.8 = 124.16

- We can look for a pattern for the total distance traveled by geometric progression is as such:

d = a*( 1 - r^n) / ( 1 - r )

Where, r = 0.8 , and a = 32

d = 32*( 1 - 0.8^n) / 0.2 = 160*( 1 - 0.8^n)

Where, n = 1 , 2 , 3 , 4 , ......

Step2247 [10]4 years ago

5 0

Answer:

If it falls from 32 feet, how could the distance be 29 feet? Twelve bounces later, 58 is "obviously" incorrect as well. Eliminate those two before you do anything else.

The total distance up until that the nth bounce is

Sn = (32 - 32(.8)12) / (1 - .8) = 149.004883722... = 149

Explanation:

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The base of a right prism is a rhombus with diagonals of 6 and 8. If the altitude of the prism is 12, what is the total surface

Agata [3.3K]

Answer:

total surface area is 432

Explanation:

Given data

base = 6

diagonals = 8

altitude = 12

to find out

total surface area

solution

we know total surface area of prism is

total surface area = lateral surface area + 2base area ..............1

first we calculate base perimeter i.e = 2 length + 2 width

so perimeter = 2(8) + 2(6) = 25

and area = length * width = 8*6 = 48

so lateral surface area is perimeter * height i.e

lateral surface area = 28* 12

lateral surface area = 336

put this value in equation 1 we get

total surface area = lateral surface area + 2base area

total surface area = 336 + 2(48)

total surface area is 432

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3 years ago

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Gemiola [76]

Answer:

Time taken, $T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

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Also, $r=l\ sin\theta$

Equation (1) becomes :

$mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}$

$v=\sqrt{gl\ tan\theta.sin\theta}$ ...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

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$T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}$

$T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}$

On solving above equation :

$T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

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