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

[Eclair_de_XII]

Here, I'll show you.

$[x_1−x_2+2.4=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64}−(t_2^2−\frac{1}{4}t_2+\frac{1}{64})]$
$x_1−x_2+2.4=(4.9)(t_1^2+\frac{1}{4}t_1−t_2^2+\frac{1}{4}t_2)$
$x_1-(x_1+\frac{1}{8})+2.4=(4.9)(t_1^2+\frac{1}{4}t_1−(t_1^2+\frac{1}{4}t_1+\frac{1}{64})+\frac{1}{4}(t_1+\frac{1}{8}))$
$2.275=(4.9)(t_1^2+\frac{1}{4}t_1−t_1^2-\frac{1}{4}t_1-\frac{1}{64}+\frac{1}{4}t_1+\frac{1}{32})$
$2.275=(4.9)(\frac{1}{4}t_1+\frac{1}{64})$
$\frac{1}{4}t_1=0.4487$
$t_1=1.795s$

 

[Eclair_de_XII]

Wait; maybe I calculated it wrong again. When I use the same value in my initial time calculation, but solve for t1, I get the right difference in time. Thanks. I'm off to bed, now.

 

[Eclair_de_XII]

$0.245=\frac{1}{4}t_1-\frac{1}{64}$
$\frac{1}{4}t_1=0.2293$
$t_1=0.917092s$

$|1.042092-0.917092|=0.125$

I'll get started on the rest of this problem tomorrow.

 

[Eclair_de_XII]

$(x_1+\frac{1}{8})$

Here, too, I made a mistake. 0.125 is actually supposed to be 1.2, bringing down the value to 1.2, again.

 

[Eclair_de_XII]

This is the progress I've made so far. I'm having trouble finding the distance from the bottom of the window to the bottom of the building.

$x_1-x_0=\frac{1}{2}(9.8\frac{m}{s^2})(1.042092s)^2$
$x_1-x_0=5.3212m$
$5.3212m=v(1.042092s)-\frac{1}{2}(9.8\frac{m}{s^2})(1.042092s)^2$
$v=10.2125\frac{m}{s}$

$v_0=10.2125\frac{m}{s}$
$x_1=5.3212m+x_0$
$x_3=0m$
$t_3=2s$

Since the time it took for the ball to pass the window going down is equal to the time for the ball to pass it going up, I will assume that the window's in the middle of the building.

$-5.3212m-x_0=(10.2125\frac{m}{s})(1s)+(4.9\frac{m}{s^2})(1s)^2$
$-x_0=(15.1125m+5.3212m)$
$x_0=-20.43m$

I'm not sure what to make of the negative sign...

 

[SteamKing]

This is the progress I've made so far. I'm having trouble finding the distance from the bottom of the window to the bottom of the building.

$x_1-x_0=\frac{1}{2}(9.8\frac{m}{s^2})(1.042092s)^2$
$x_1-x_0=5.3212m$
$5.3212m=v(1.042092s)-\frac{1}{2}(9.8\frac{m}{s^2})(1.042092s)^2$
$v=10.2125\frac{m}{s}$

$v_0=10.2125\frac{m}{s}$
$x_1=5.3212m+x_0$
$x_3=0m$
$t_3=2s$

Since the time it took for the ball to pass the window going down is equal to the time for the ball to pass it going up, I will assume that the window's in the middle of the building.

$-5.3212m-x_0=(10.2125\frac{m}{s})(1s)+(4.9\frac{m}{s^2})(1s)^2$
$-x_0=(15.1125m+5.3212m)$
$x_0=-20.43m$

I'm not sure what to make of the negative sign...

I think you're trying too hard to use a negative sign in your calculations when you don't need one.

If we take x0 as the top of the building, we can arbitrarily set x0 = 0 m and make that our reference point. We can also say that we drop the ball at t = 0 sec. and use that as a reference point for our timing of the fall of the ball.

For each second after the ball is dropped, the ball travels so many meters, which distance we add to our reference point for distance, x0 = 0 m.

For example, when t = 1.042 sec., the ball has dropped 5.321 m approx., which is where it passes the bottom of the window on its way to the ground. We are told the ball spends a total of 2.0 seconds below the bottom of the window while it falls to the ground, hits, and then bounces back up to the bottom of the window. From this description, we can infer that the total time it took the ball to fall from the roof to the ground is 1.042 + 1.0 = 2.042 sec. Since the initial velocity of the ball was 0 m/s, it becomes very easy to calculate the height of the building using the total time of free fall.

In these situations, making a simple sketch helps to clarify things.

 

[Eclair_de_XII]

$x=x_0+v_0t+\frac{1}{2}at^2$
$x=\frac{1}{2}(9.8\frac{m}{s^2})(2.042s)^2$
$x=20.43m$