CPhill

The horizontal velocity is unaffected by the drop..   also ....the weight of the marble is irrelevant..so....

The time it takes to drop  =  d / r  =    .51 m / 1.2 m/s   =  .425 s

So......the height of the table can be found  as

H  =  vi * t   +  (1/2) a  * t^2

Where  

H  =  the table height

vi =  the intital vertical velocity =  0

t  =  time to drop  =   .425 s

a  =  acceleration due to gravity =  9.81 m/s^2

So we have

H  =  0 * .425s +  (1/2) (9.81m/s^2) * (.425s)^2  

H  ≈   .886 m

We can use this equation

h  = vi * t   +  (1/2) * a * t^2

Where

h = height  =  8.5 m  

vi  =  initial velocity  = 0

a  = acceleration  due to gravity   

t  = time  =  5s

8.5 m  =  0 * 5s    +  (1/2) * a  * (5 s)^2

8.5  =  12.5  a

8.5 m / 12.5 s^2    =   .68 m/s^2 =  acceleration due to gravity

This  will  be  a negative quantity since it is directed downward

So..... this is      -.68  / -9.81    ≈  .069  that of Earth  ≈  6.9%  of that on Earth

Thanks, Donut  and Jeff......!!!

-1115  / 134   ≈  -8.32  ft / m   = - [ 8 + 32/100 ]  =  - [ 8 + 4/25  ]  ft / m

We need  this formula

vf^2  =  vi^2  +  2 (a)(d)

Where  

vi  is the initial velocity  = 0

a  =  acceleration due to gravity  =  -9/81m/s^2

d  = the displacement  =  321 m

vf   =  the final  velocity  ......so we have....

vf^2  = 0  +  2 (-9.81m/s^2)(321m)

vf^2  =   6298.02 m^2/s^2           take the square root

vf  ≈  79.36 m/s

I'm not great at Physics....but....I might be able to answer a few of these

Acceleration  =   Change  in Velocity /  Change in Time =   [ 30 - 0 ] / [ 0 - 4.2 ] =

[ 30m/s]  / [ - 4.2 s]  ≈   - 7.14 m/s^2

Here are the constraints

x  + y  ≤ 36

y  ≥  3x

And we seek to maximize this

2x  +  3y

Look  at the graph of the constraints :  https://www.desmos.com/calculator/ezk3fnwmsj

The max occurs at the only corner point,  (9,27)

So......the max profit  is

2(9)  + 3(27)  =

18  +  81   =

$99

Thanks, Asad....!!!

1. (a^2-6)^2 - (4a + 6)^2

This is a difference of squares

[ (a^2 - 6) + (4a + 6)]  [ [(a^2 - 6) - (4a + 6) ]  =

[ a^2 + 4a]  [ a^2 - 4a - 12 ]  =

[a] [a + 4 ] [a - 6] [a + 2]

2. (2m -3)^2 - (8m^2 - 4)^2

Same thing

[ (2m - 3) + (8m^2 - 4) ]  [(2m - 3) - (8m^2 - 4) ]  =

[ 8m^2 + 2m - 7 ]  [ -8m^2 + 2m + 1 ]  =

[8m^2 + 2m - 7 ] [-1] [8m^2 - 2m - 1]  =

[8m^2 + 2m - 7 ] [-1] [ 2m - 1] [ 4m + 1 ] =

- [8m^2 + 2m - 7 ]   [ 2m - 1 ] [4m + 1]  

3. (17p^2 -26) - (8p^2 -4)^2

(17p^2 - 26) -  [ (8p^2 - 4) (8p^2 - 4) ] =

( 17p^2  - 26)  -  [ 64p^4 - 64p^2 + 16 ]  =

 [ - 64p^4 + 64p^2 + 17p^2  - 42 ]  =

 -64p^4  +  81p^2  - 42    ⇒    this is irreducible

Let one of the parallel sides   = x......so the other is 22-x

And  the length of the  midline  = sum of the parallel sides / 2  =  22/2  = 11

Let the area of the smaller part  =  (h/2)(11 + 22 -x)  = (h/2)(33 - x)

Let the area of the larger part  = (h/2)(11 + x) 

And  (7/4)*area  of the smaller part   = the larger part....so.....

(7/4)*(h/2)*(33-x)  = (h/2)*(11+ x)

(7/4) (33 - x )   =  (11 + x)

7(33 - x)  = 4(11 + x)

231  - 7x   =  44 + 4x       add 7x to both sides, subtract 44 from both sides

187  =  11x         divide both sides by 11

17  = x

So.....one parallel base  = 17  and the other is  22-17  = 5

And their product  = 17 * 5   =  85