# Augustine in septic tank (Alice in Wonderland)

1. Dec 5, 2011

### maxtheminawes

1. The problem statement, all variables and given/known data
Flopping about in pain, Augustine starts rolling across the beach at a constant 1.4m/s. This isn’t Augustine’s day as he now rolls into a hole somebody had just dug for a new septic tank. If it takes him 1.01s to hit the bottom of the hole:
a)How deep is the hole?
b)what is hid acc as he falls?
c)what is his final displacement?

2. Relevant equations
y=yo+vot+1/2at^2
x=xo+vot+1/2at^2
yo=initial y displacement
vo=initial velocity
xo=initial x displacement

3. The attempt at a solution
a)y=yo+vot+1/2at^2
y=1.4sin270+1/2(-9.8)(1.01^2)
y=-6.4
y=6.4m

b)-9.8m/s^2

c)x=xo+vot+1/2at^2
x=1.4(1.01)
x=1.41m
sqrt of (1.41^2+6.4^2)=6.55m
tan-1(6.4/1.41)=77.58
6.55m @ 77.58degrees
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 6, 2011

### Stonebridge

a) the vertical falling motion is independent of the horizontal, so there is no need to include the horizontal velocity in this. (Where does the angle 270 come from, by the way?)
The initial vertical velocity is zero in this formula.
b) correct
c) you need to use the correct answer for a) for the vertical displacement. The horizontal displacement is correct.

3. Dec 6, 2011

### maxtheminawes

ok, i redid the problem
a)y=1/2(-9.8)(1.01^2)
y=5m
c)x=1.41
y=5
sqrt of (1.41^2+5^2)=5.2m

is this correct?

4. Dec 7, 2011

### Stonebridge

Yes that's fine now.