# How do I find distance given time and an arbitrary distance?

1. Feb 27, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"A ball is dropped from a building's roof and passes a window, taking 0.125 s to fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.125 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.00 s. How tall is the building?"

2. Relevant equations
$x_0=0m$ (top of building)
$x_1=1.2m$
$t_1=t_2=0.125s$
$t_3=2s$
$v_0=0\frac{m}{s}$
$a=9.8\frac{m}{s^2}$
Find:
$v_1$
$v_2$
for
$x_2$ (bottom of building)

$v=v_0+at$
$x-x_0=vt-at^2$
$v^2=v_0^2+2a(x-x_0)$

3. The attempt at a solution
$1.2m=v_1(0.125s)-(4.9\frac{m}{s^2})(0.125s)^2$
$v_1=10.2125\frac{m}{s}$
$v_2=10.2125\frac{m}{s}+(-9.8\frac{m}{s^2})(1.875s)=-8.1625\frac{m}{s}$
$(-8.1625\frac{m}{s})^2-(10.2125\frac{m}{s})^2=(-8.1625\frac{m}{s^2}-10.2125\frac{m}{s^2})(x_2-1.2m)$
$-37.67\frac{m^2}{s^2}=-18.375\frac{m}{s^2}(x_2-1.2m)$
$x_2=3.25m$

Last edited: Feb 27, 2016
2. Feb 27, 2016

### SteamKing

Staff Emeritus
Always check the formatting of your post by hitting the Preview button on the Lower RHS of the edit box.

The quantities used in your work are not clear. How tall is the building?

3. Feb 27, 2016

### Eclair_de_XII

It's supposed to be 20.4 m, but I ended up with 3.25 m.

4. Feb 27, 2016

### SteamKing

Staff Emeritus
Although the height of the window is given as 1.2 m, you are not told exactly where the top or the bottom of the window is located along the side of the building.
You must work this out from the free fall times given in the problem statement.

It looks like x0 is supposed to represent the top of the building, but the distance from x0 to the top of the window must be worked out. Use the SUVAT equations to write an expression for the time it takes for the ball to pass the window, namely 0.125 s, which is 1.2 m tall.

5. Feb 27, 2016

### JustDerek

I'm not 100% sure as I'm just learning this stuff myself. But I think you can work out the velocity to the bottom of the window by just doing $v_1=d/t$ calling this velocity $v_1$ then put this in as an initial velocity in your $v=v_0+at$ to find out your v over the two seconds. Then use that to workout distance eventually. Sorry if this is wrong but it's how I'd approach it

6. Feb 27, 2016

### Eclair_de_XII

So wait... The window is not 1.2 m from the top; it's 1.2 m tall?

7. Feb 27, 2016

### SteamKing

Staff Emeritus
No, v ≠ d / t here becuz the ball is accelerating as it free falls. The velocity is not constant.

8. Feb 27, 2016

### SteamKing

Staff Emeritus
Correct. You can check this by re-reading the problem statement carefully.

What you are given is the time it takes for the falling ball to pass the window, namely 0.125 sec. Since the ball is free falling from the roof, you can use the SUVAT equations to write an expression for the time it takes the ball to fall from the roof to the top of the window, and another expression for the time it takes the ball to fall from the roof to the bottom of the window. The difference in these two times must equal 0.125 s.

9. Feb 27, 2016

### JustDerek

Ah yeah. Good point. I won't try and input anymore I've already been wrong once

10. Feb 28, 2016

### Eclair_de_XII

$t=\frac{1}{8}s$
$a=9.8\frac{m}{s^2}$
$x=0m$
$x_0=1.2m$

I'm going to be doing initial and final velocity at the same time, if that doesn't change anything.

$x-x_0=v_*t±\frac{1}{2}at^2$
$-1.2m=v_*(\frac{1}{8}s)±\frac{1}{2}(9.8\frac{m}{s^2})(\frac{1}{8}s)^2$
$-1.2m±(4.9\frac{m}{s^2})(\frac{1}{64}s^2)=v_*(\frac{1}{8}s)$
$-1.2m±0.0765625m=v_*(\frac{1}{8}s)$
$v_*=-9.6±0.6125m$
$v_0=-10.2125\frac{m}{s}$
$v=-8.9875\frac{m}{s}$

Now I'm confused, because the magnitude of the final velocity shadows that of the initial velocity...

Irrelevant, but please do not mock me.

11. Feb 28, 2016

### SteamKing

Staff Emeritus
Working with velocities here is not an efficient method of solution, especially since you want to find the overall height of the building.

You are given time differentials to work with and distances to solve for, so I recommend that you stick with the SUVAT formulas which involve just time and distance.

12. Feb 28, 2016

### Eclair_de_XII

There's not really a SUVAT equation that excludes both initial and final velocities, is there?

13. Feb 28, 2016

### SteamKing

Staff Emeritus
For this case, where the ball is free falling with an initial velocity of 0 m/s, I think you can find a SUVAT equation which involves time and distance without any velocity calculation.

14. Feb 28, 2016

### cnh1995

You have s=1.20m, t=0.125s and a=10m/s2... You can find the downward velocity 'u' of the ball when it is at the top of the window(using s=ut+at2/2) and velocity 'v' of the ball when it is at the bottom of the window. Rest of the calculations will be straightforward.

15. Feb 28, 2016

### Eclair_de_XII

But it's not really 0 m/s from the time it reaches the top of the window, is it?

16. Feb 28, 2016

### SteamKing

Staff Emeritus
I never said it was.

The initial velocity of the ball, when it is dropped from the roof of the building, is 0 m/s.

Reference all of the distances from the roof of the building to find the free fall times to the top of the window and the bottom of the window. You'll come up with two equations, one each expressing the distance to the top and bottom of the window.

17. Feb 28, 2016

### Eclair_de_XII

Since I don't know where the window is, this is honestly as far as I can go.

$x-x_0=v_0t+\frac{1}{2}at^2$
$x_0=0m$
$x=?$
$x=\frac{1}{2}(9.8\frac{m}{s^2})t$
$x=4.9\frac{m}{s^2}(t)$

18. Feb 28, 2016

### cnh1995

Once you get the velocities of the ball at the top and bottom of the window(downward motion), you can calculate the total height of the building.

19. Feb 28, 2016

### Eclair_de_XII

But I don't believe that the final velocity can be less in magnitude than the initial velocity, from the point that the ball passes the top of the window.

20. Feb 28, 2016

### cnh1995

For downward motion, acceleration will be positive.