# Homework Help: Motion due to gravity

1. Sep 6, 2010

### furor celtica

1. The problem statement, all variables and given/known data

When an object falls through a liquid, three forces act on it: its weight, the buoyancy and the resistance of the liquid. Two spheres, of mass 0.5 kg and 1.5 kg respectively, have the same radius, so that they have the same buoyancy of 3.2 newtons, and the same resistance formula, 5v newtons when falling at speed v ms^-1. Both spheres enter the liquid falling vertically at 1 ms^-1. Calculate the terminal speeds of the two spheres, and the acceleration or deceleration when they enter the liquid.
(Here I found: for mass 0.5 kg, terminal speed 0.36 ms^1 and acceleration -6.4 ms^2; for mass 1.5 kg, terminal speed 2.36, acceleration 4.53 (to 3 s.f.))
If the depth of the liquid is 10m, show that the heavier sphere reaches the bottom after a time between 4 and 10 seconds. Find bounds for the time that the lighter sphere takes to reach the bottom.

2. Relevant equations

Ok to starting with the heavier sphere I thought it would be logical to proceed like this:
To find higher bound: distance = 10, u=1, v<2.36, a= (15 – 3.2 – 5v)/1.5 = (11.8 -5v)/1.5, equation s=ut + 0.5at^2, find v in terms of t and find interval for which v<2.36 is true

3. The attempt at a solution

No problem here, I found t<10
I used a similar method to find the lower bound using s=vt-0.5at^2 and v=1, u>1, but this time the expression had no real roots, so I’m wondering if I’m not using the wrong method altogether, and if my first result was just lucky.
Can someone show me how to work through this?

2. Sep 6, 2010

### Delphi51

Welcome to PF, furor!

I think that lower bound on the time is just how long it takes if it is assumed it goes at the terminal velocity for the whole 10 meters.

Your a= (15 – 3.2 – 5v)/1.5 from the F = ma law looks good, though I didn't get a nice even 15.
Anyway, the question is how you can find the terminal velocity with this. What is the acceleration when the terminal velocity is reached? Plug that in, and you've got it!

Careful with s=ut + 0.5at^2; must keep in mind that the acceleration is not constant so this formula doesn't apply, except perhaps in estimating an upper or lower bound.

3. Sep 7, 2010

### furor celtica

but i already have the terminal velocity, i need the time interval.
how can i find this without the constant acceleration formulae? if i can't use s=ut + 0.5at^2 then all my working is useless anyway, including my apprently correct t<10 result.
can you please explain more, thanks for your help btw this is the only forum where somebody has tried to help me with this question

4. Sep 7, 2010

### rl.bhat

(Here I found: for mass 0.5 kg, terminal speed 0.36 ms^1 and acceleration -6.4 ms^2; for mass 1.5 kg, terminal speed 2.36, acceleration 4.53 (to 3 s.f.))

Will you please show me how you calculated the terminal velocity and acceleration?
When the balls enter the water, there initial velocity before entering the water is 1 m/s and there acceleration is g. Inside the water there velocity increases with variable acceleration until they attain the terminal velocity. Duration to attain the terminal velocity by the two spheres is different and hence the time taken by them to reach the bottom is different.
And can you explain this?
If the depth of the liquid is 10m, show that the heavier sphere reaches the bottom after a time between 4 and 10 seconds.
Do they mean 4 s to reach the terminal velocity and 10 s to reach the bottom?

5. Sep 7, 2010

### furor celtica

i dont know what they mean by this exactly, maybe they mean find the times taking u=1 and v=terminal speed, i really dont know, thats what i took it to be
about the accelerations and terminal velocities i checked them already, its all good

6. Sep 7, 2010

### rl.bhat

I can understand the calculation of the terminal velocities. But I am unable to understand the positive and negative acceleration.

7. Sep 7, 2010

### Delphi51

So the speed is > 2.36. Therefore the time to fall is < 10/2.36 = 4.2 s or approximately 4 s since this is just an estimate.
Also you have that the average speed is < 1 so time to fall is > 10/1. This is as good as you can do without the formulas for motion with varying acceleration - have to save that for next year I think!

BTW, rounding g*1.5 = 14.7 to 15 has left the calcs with only 2 digit accuracy.

8. Sep 8, 2010

### furor celtica

ok i get it now its al ot simpler than i thought, just using constant velocity over 10 m with values u=1 and v=terminal velocity, am i right?