Augustine in septic tank (Alice in Wonderland)

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Homework Help Overview

The problem involves a scenario where Augustine rolls into a hole while moving horizontally at a constant speed. The discussion focuses on calculating the depth of the hole, the acceleration during the fall, and the final displacement, incorporating both vertical and horizontal components of motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the independence of vertical and horizontal motions, question the initial vertical velocity, and discuss the correct application of kinematic equations for both vertical and horizontal displacements.

Discussion Status

Some participants have provided corrections and alternative calculations for the depth of the hole and the final displacement. There is an acknowledgment of the revised calculations, but no explicit consensus on the final correctness of the results.

Contextual Notes

Participants are working within the constraints of the problem statement and the equations provided, while also questioning assumptions about initial conditions and the application of kinematic principles.

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Homework Statement


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?



Homework Equations


y=yo+volt+1/2at^2
x=xo+volt+1/2at^2
yo=initial y displacement
vo=initial velocity
xo=initial x displacement



The Attempt at a Solution


a)y=yo+volt+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+volt+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
 
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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.
 
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?
 
maxtheminawes said:
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?

Yes that's fine now.
 

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