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

[Eclair_de_XII]

Homework Statement

"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?"

Homework 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)

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

The Attempt at a Solution

$1.2m=v_1(0.125s)-(4.9\frac{m}{s^2})(0.125s)^2$
$v_1=10.2125\frac{m}{s}$
$v_2=10.2125\frac{m}{s}+(-9.8\frac{m}{s^2})(1.875s)=-8.1625\frac{m}{s}$
$(-8.1625\frac{m}{s})^2-(10.2125\frac{m}{s})^2=(-8.1625\frac{m}{s^2}-10.2125\frac{m}{s^2})(x_2-1.2m)$
$-37.67\frac{m^2}{s^2}=-18.375\frac{m}{s^2}(x_2-1.2m)$
$x_2=3.25m$

 

[SteamKing]

Homework Statement

"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 a nexact reverse of the fall. The time the ball spends below the bottom of the window is 2.00 s. How tall is the building?"

Homework 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)

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

The Attempt at a Solution

$1.2m=v_1(0.125s)-(4.9\frac{m}{s^2})(0.125s)^2$
$v_1=10.2125\frac{m}{s}$
$v_2=10.2125\frac{m}{s}+(-9.8\frac{m}{s^2})(1.875s)=-8.1625\frac{m}{s}$
$(-8.1625\frac{m}{s})^2-(10.2125\frac{m}{s})^2=(-8.1625\frac{m}{s}-10.2125\frac{m}{s})(x_2-1.2m)$
$-37.67\frac{m^2}{s^2}=-18.375\frac{m}{s}(x_2-1.2m)$
$x_2=3.25m$

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?

 

[Eclair_de_XII]

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

 

[SteamKing]

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

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.

 

[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

 

[Eclair_de_XII]

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.

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

 

[SteamKing]

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

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

 

[SteamKing]

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

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.

 

[JustDerek]

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

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

 

[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...

No, v ≠ d / t here becuz the ball is accelerating as it free falls.

Irrelevant, but please do not mock me.

 

[SteamKing]

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

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...

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.

 

[Eclair_de_XII]

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

 

[SteamKing]

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

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.

 

[cnh1995]

taking 0.125 s to fall from the top to the bottom of the window, a distance of 1.20 m.

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.

 

[Eclair_de_XII]

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

 

[SteamKing]

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

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.

 

[Eclair_de_XII]

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

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

 

[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.

 

[Eclair_de_XII]

$v_*=-9.6±0.6125m$
$v_0=-10.2125\frac{m}{s}$
$v=−8.9875\frac{m}{s}$

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.

 

[cnh1995]

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.

For downward motion, acceleration will be positive.

 

[SteamKing]

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

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

It's better to assume the top of the window is located at x1 and the bottom is at x2, measured from the roof. You can assume it takes the ball t1 to get to x1 and t2 to get to x2. You know what x2 - x1 is and what t2 - t1 is. This information will locate the window from the roof of the building, which is the first part of calculating how tall the building is.

 

[Eclair_de_XII]

I actually did use a positive acceleration.

$-1.2m=v_*(\frac{1}{8}s)±\frac{1}{2}(9.8\frac{m}{s^2})(\frac{1}{8}s)^2$
$-1.2m±(4.9\frac{m}{s^2})(\frac{1}{64}s^2)=v_*(\frac{1}{8}s)$
$−1.2m±0.0765625m=v_∗(\frac{1}{8}s)$
$v_∗=−9.6±0.6125\frac{m}{s}$

 

[cnh1995]

1.2=v_topt+4.9t²...
Once you get v_top, you can find v_bottom=v_top+9.8t..

 

[Eclair_de_XII]

Well, I tried following King's advice, and ended up with this. I will completely understand you if you say this doesn't make any sense.

$x_2-x_1=1.2m$
$x_2=x_1+1.2m$
$x_1=x_2-1.2m$
$t_2-t_1=\frac{1}{8}s$
$t_2=t_1+\frac{1}{8}s$
$t_1=t_2-\frac{1}{8}s$
$v_0=0\frac{m}{s}$
$t_0=0s$

I'm going to eliminate the units, because it's confusing. Nevertheless, I still end up with a nonsensical answer. I basically subtracted the equations from each other.

$x_1+1.2-x_0=(4.9)(t_1+\frac{1}{8}s)^2$
$x_2-1.2-x_0=(4.9)(t_2-\frac{1}{8}s)^2$
$x_1+1.2-x_0=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64})$
$x_2-1.2-x_0=(4.9)(t_2^2-\frac{1}{4}t_1+\frac{1}{64})$
$[x_1+1.2-x_0=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64})]$
$-[x_2-1.2-x_0=(4.9)(t_2^2-\frac{1}{4}t_1+\frac{1}{64})]$
$=[x_1-x_2-2.4=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64}-(t_2^2-\frac{1}{4}t_2+\frac{1}{64})]$
$x_1-x_2-2.4=(4.9)(t_1^2+\frac{1}{4}t_1-t_2^2+\frac{1}{4}t_2)$

Substitute...

$(x_2-1.2)-x_2-2.4=(4.9)[(t_2^2-\frac{1}{4}t_2+\frac{1}{64})+\frac{1}{4}(t_2-\frac{1}{8})-t_2^2+\frac{1}{4}t_2)]$
$\frac{-3.6}{4.9}=t_2^2-\frac{1}{4}t_2+\frac{1}{64}+\frac{1}{4}t_2-\frac{1}{32}-t_2^2+\frac{1}{4}t_2)$
$-0.7347=\frac{1}{4}t_2-\frac{1}{64}$
$\frac{1}{4}t_2=-0.72s$
$t_2=-2.88s$

And I end up with a negative time. I'm definitely doing something wrong, but I don't know what. And when I did it on paper, I still had seconds-squared.

 

[cnh1995]

Well, I tried following King's advice, and ended up with this. I will completely understand you if you say this doesn't make any sense.

$x_2-x_1=1.2m$
$x_2=x_1+1.2m$
$x_1=x_2-1.2m$
$t_2-t_1=\frac{1}{8}s$
$t_2=t_1+\frac{1}{8}s$
$t_1=t_2-\frac{1}{8}s$
$v_0=0\frac{m}{s}$
$t_0=0s$

I'm going to eliminate the units, because it's confusing. Nevertheless, I still end up with a nonsensical answer. I basically subtracted the equations from each other.

$x_1+1.2-x_0=(4.9)(t_1+\frac{1}{8}s)^2$
$x_2-1.2-x_0=(4.9)(t_2-\frac{1}{8}s)^2$
$x_1+1.2-x_0=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64})$
$x_2-1.2-x_0=(4.9)(t_2^2-\frac{1}{4}t_1+\frac{1}{64})$
$[x_1+1.2-x_0=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64})]$
$-[x_2-1.2-x_0=(4.9)(t_2^2-\frac{1}{4}t_1+\frac{1}{64})]$
$=[x_1-x_2-2.4=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64}-(t_2^2-\frac{1}{4}t_2+\frac{1}{64})]$
$x_1-x_2-2.4=(4.9)(t_1^2+\frac{1}{4}t_1-t_2^2+\frac{1}{4}t_2)$

Substitute...

$(x_2-1.2)-x_2-2.4=(4.9)[(t_2^2-\frac{1}{4}t_2+\frac{1}{64})+\frac{1}{4}(t_2-\frac{1}{8})-t_2^2+\frac{1}{4}t_2)]$
$\frac{-3.6}{4.9}=t_2^2-\frac{1}{4}t_2+\frac{1}{64}+\frac{1}{4}t_2-\frac{1}{32}-t_2^2+\frac{1}{4}t_2)$
$-0.7347=\frac{1}{4}t_2-\frac{1}{64}$
$\frac{1}{4}t_2=-0.72s$
$t_2=-2.88s$

And I end up with a negative time. I'm definitely doing something wrong, but I don't know what. And when I did it on paper, I still had seconds-squared.

You can calculate v_top and v_bottom as I mentioned in #23. Then using v_bottom and time t=1s, you can calculate height below the window, and using v_top, you can calculate the height above the window. Just add them with window height to get the total height.

 

[Eclair_de_XII]

Wait, I think I found the error in my operations.

$[x_1+1.2-x_0=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64})]$
$−[x_2-1.2-x_0=(4.9)(t_2^2-\frac{1}{4}t_1+\frac{1}{64})]$
$=[x_1−x_2+2.4=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64}−(t_2^2−\frac{1}{4}t_2+\frac{1}{64})]$

$(x_2-1.2)-x_2+2.4=(4.9)[(t_2^2-\frac{1}{4}t_2+\frac{1}{64})+\frac{1}{4}(t_2-\frac{1}{8})-t_2^2+\frac{1}{4}t_2)]$
$(\frac{1.2}{4.9})=t_2^2-\frac{1}{4}t_2+\frac{1}{64}+\frac{1}{4}t_2-\frac{1}{32}-t_2^2+\frac{1}{4}t_2)$
$0.245=\frac{1}{4}t_2-\frac{1}{64}$
$\frac{1}{4}t_2=0.2605$
$t_2=1.042092s$

 

[SteamKing]

Wait, I think I found the error in my operations.

$[x_1+1.2-x_0=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64})]$
$−[x_2-1.2-x_0=(4.9)(t_2^2-\frac{1}{4}t_1+\frac{1}{64})]$
$=[x_1−x_2+2.4=(4.9)(t_1^2+\frac{1}{4}t_1+\frac{1}{64}−(t_2^2−\frac{1}{4}t_2+\frac{1}{64})]$

$(x_2-1.2)-x_2+2.4=(4.9)[(t_2^2-\frac{1}{4}t_2+\frac{1}{64})+\frac{1}{4}(t_2-\frac{1}{8})-t_2^2+\frac{1}{4}t_2)]$
$(\frac{1.2}{4.9})=t_2^2-\frac{1}{4}t_2+\frac{1}{64}+\frac{1}{4}t_2-\frac{1}{32}-t_2^2+\frac{1}{4}t_2)$
$0.245=\frac{1}{4}t_2-\frac{1}{64}$
$\frac{1}{4}t_2=0.2605$
$t_2=1.042092s$

Knowing t1 or t2 should allow you to calculate how far the top or bottom of the window is from the roof.

Now, the second part of finding the height of this building involves what happens to the ball after it passes below the bottom of the window.
You're told the ball spends a total of 2 sec. falling, hitting the ground, and bouncing back to the window. You are also told that the upward flight of the ball is the exact reverse of the fall below the window.

Can you take your previous calculations and find out how high the bottom of the window is above the ground?

 

[Eclair_de_XII]

Sorry; it's late at night, and I'm tired. I'll try next morning.

 

[Eclair_de_XII]

And I just realized my time was wrong; $|t_2-t_1|≠0.125$

 

[SteamKing]

And I just realized my time was wrong; $|t_2-t_1|≠0.125$

Where did you calculate an actual value for t1? I didn't see one in your calculations.

 

[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]

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