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

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x