If a coin is thrown with a starting velocity of 0 m/s down a dry well and hits bottom in 1.2 s, what's the depth of the well? with what velocity does the coin in the previous problem hit the bottom?

Depth = 7.06 m

Velocity = 11.77 ms-1

Step-by-step explanation:

We use the equation of motion:-

s = ut + 1/2 g   where s = distance fallen , t = time , u = initial velocity and g = 9.81 ms^-2 , the acceleration due to gravity.

So s = 0(t) + 1/2 * 9.81 (1.2)^2

= 7.06 m to nearest tenth.

We use another equation of motion to find the final velocity:-

v^2 = u^2 + 2gs    where v = final velocity

v^2 = 0 + 2*9.81 * 7.06

= 138.52

v = √138.52 = 11.77 m s-1

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: