Stupidly basic physics/math question

Enlightenment · Post by **Enlightenment** » 2003-01-23 11:48pm

I know this is probably dead simple but I've been tearing my hair out trying to figure this out. If I shoot mass_1 directly upwards at 10m/s, wait one second, then launch mass_2 directly upwards at 25m/s, at what altitude (or at what time) will mass_1 be at the same height as mass_2? Under Earth gravity.

I can get as far as

y_mass1 = 10t + 1/2gt^2

y_mass2 = 25(t-1) / 1/2g(t-1)^2

10t + 1/2gt^2 = 25(t-1) / 1/2g(t-1)^2

but at this point I can't figure out how to solve for t. Can someone here point me in the right direction?

Wicked Pilot · Post by **Wicked Pilot** » 2003-01-24 12:02am

try this, plot the two on a graph, with your x axis being time, and your y axis being height. Where the two lines meet is your answer.

Enlightenment · Post by **Enlightenment** » 2003-01-24 12:03am

That's cheating.

I'm trying to do this analytically.

Master of Ossus · Post by **Master of Ossus** » 2003-01-24 12:06am

You cannot solve for "t" in the manner that you are working it. You first must set the two equations up so that Mass 1 has t=0 when Mass 2 has t=0. To do this, add a constant to Mass 1's altitude, keeping in mind the acceleration. Basically, calculate Mass 1's altitude, velocity, and acceleration for t=1. Then add that altitude to it, and reset it so that t=0 for both of them. You then have two equations, and two unknowns, and you can easily solve for the intersect between the two.

Master of Ossus · Post by **Master of Ossus** » 2003-01-24 12:08am

Set it up like this:

y mass 1= 10(t+1)+(1/2)g(t+1)^2

y mass 1=10t+10+ (1/2)g(t^2+2t+1)

y mass 1=10t+10+ (t^2/s+t+1/s)g

Wicked Pilot · Post by **Wicked Pilot** » 2003-01-24 12:15am

Equation for Mass 1: x=10t+(1/2)(-9.8)t^2
Equation for Mass 2: x=25(t-1)+(1/2)(-9.8)(t-1)^2

x=height, t=time

10t+(1/2)(-9.8)t^2=25(t-1)+(1/2)(9.8)(t-1)^2

solve for t, and that is your time

plug t into either two initial equations, and you get the height of the collision.

time=1.2s
height=4.9m

Enlightenment · Post by **Enlightenment** » 2003-01-24 02:23am

Wicked Pilot wrote: 10t+(1/2)(-9.8)t^2=25(t-1)+(1/2)(9.8)(t-1)^2

Errr, that's basically what I posted. I can get that far. How can I isolate t from that mess, however? I can't use the quadratic formula or a root solver because neither equation should be equated to zero. I can't use the negative of either equation as the first term of the other and apply the quadratic equation because neither equation can be reduced to a value without first solving the other.

If I use MoO's suggestion then I wind up with

X_1L + 10t + 1/2gt^2 = 25t + 1/2gt^2

where X_1L is the initial altitude of mass_1 (5.1m). This is simpler but still leaves me with the same problem of how to equate two quadratic equations.

Enlightenment · Post by **Enlightenment** » 2003-01-24 02:57am

I've taken another shot at this and I get:

(X_bl is the height of mass 1 (_b) after one second)

X_bl + V_b t + 1/2gt^2 = V_s t + 1/2gt^2

X_bl + V_b t + 1/2gt^2 - V_s t - 1/2gt^2 = 0

The gt^2 terms cancel, therefore:

X_bl + V_b t - V_s t = 0

t (V_b - V_s) = -X_bl

t = -X_bl / (V_b - V_s)

Plugging in the numbers

-X_bl = 5.1
V_b = 10m/s
V_s = 25m/s

t = .43s after v_b has been in the air for 1s

t = 1s + .43s = 1.43s.

This is fairly close to Wicked Pilot's answer but not close enough. Have I made any obvious mistakes? As you can probably tell I know just enough math to be dangerous but not enough to get it right...

Cyborg Stan · Post by **Cyborg Stan** » 2003-01-24 03:23am

It could be because I'm tired, but I don't see the problem. Write out all the terms on both sides of the equations, and you should get a .5gt² on both sides that would cancel out. Solving for t should be a snap then.

Cyborg Stan · Post by **Cyborg Stan** » 2003-01-24 03:31am

I see a problem. You forgot to adjust the velocity of mass 1 when you re-wrote the equation. It should be 10 m/s - (9.8m/s² * 1), or .2 m/s. Correct for that and you should get the answer Wicked Pilot and myself got. Although personally, it would've made more sense to me to simply solve it directly from time....

10t - .5gt² = 25(t-1) - .5g(t-1)²
10t - .5gt² = 25t - 25 - .5gt² + gt - .5g
25 + .5g = 15t + gt
(25 + .5g) / (15 + g) = t

Enlightenment · Post by **Enlightenment** » 2003-01-24 03:56am

Ah. That fixes a few things.

t = 1.312s (or .312s after the launch of s) with an interception altitude of 7.56m.

Thanks everyone.

Welcome to SD, Cyborg Stan. Enjoy your stay and try not to lose your sanity.

Wicked Pilot · Post by **Wicked Pilot** » 2003-01-24 10:07am

Enlightenment wrote:t = 1.312s (or .312s after the launch of s) with an interception altitude of 7.56m.

No, you're still wrong, the interception occurs at 1.2056s, at 4.934m. The first projectile goes no higher than 5.102m.

Captain Hornblower · Post by **Captain Hornblower** » 2003-01-24 11:57am

The answer is 1.34s and here is how I solved it, using kinematics.

V1o = 10m/s
V2o = 25m/s

t1 = T
t2 = T-1

H1 = H2

H = Ho + 1/2*(Vo + V)*t, where Ho1 = Ho2 = 0
V = Vo + at

We want to solve for T, therefore

H1 = 1/2 *(10 + V1)*T, where V1 = 10-9.8T
H1 = 1/2*(10 +10 - 9.8T)*T
H1 = .5*(20-9.8T)*T
H1 = 10T-4.9T^2

Similarly,
H2 = .5*(25+V2)*(T-1)
doing a little algebra gives
H2 = 25T-4.9T^2 -20.1

Setting H1 = H2 gives

10T-4.9T^2 = 25T -4.9t^2 -20.1
Cancelling terms and solving for T gives us

T = 20.1/15 = 1.34sec

Substituting backing to our height equation gives H1 = H2 = 4.6m.

Now what does all this say?

Well mass one is on its way back down when it and mass two are the same height. The max height acheived by mass one is 5.1m before it starts falling back to earth. It takes it just 1.02 seconds for mass one to get to 5.1m. In the .02 seconds mass two has been in the air it has only reached a height of .5m. Finally, there is one other solution, that is the time both masses are the same height of zero when both are back on the ground. But that is an easy and uninteresting solution.

Cyborg Stan · Post by **Cyborg Stan** » 2003-01-24 03:30pm

I really can't tell what you were trying to do with "H2 = .5*(25+V2)*(T-1)" It looks like you're saying that the height of the second mass is half of the initial velocity of the second mass plus an additional 25 m/s launched one second after the first mass, which really doesn't make a hell of alot of sense given the problem, even if you are ignoring gravity. "H2 = 25T-4.9T^2 -20.1 " makes sense, although I think you left out a 9.8t term in there. [(a+b)² = a² + 2ab + b²]

Wicked Pilot · Post by **Wicked Pilot** » 2003-01-24 05:10pm

Captain Hornblower wrote:The answer is 1.34s and here is how I solved it, using kinematics.

No, that is wrong. At 1.34s, mass 1 is at 4.6016 meters, and mass 2 is at 7.9446 meters. I don't care how you solved it, you are still wrong. Do a more careful job of checking your work next time.

Enlightenment · Post by **Enlightenment** » 2003-01-24 11:43pm

I redid this (again) using a spreadsheet for the basic maths and the answers I get are t=0.205s, height=4.93m, which is what Wicked Pilot posted the first time. I've crosschecked both figures (x_b == x_s) and they're definitely correct. I must have made some data entry errors to wind up with 1.34s.

Believe it or not at one point I was going to be an engineer. Anyone who's read this thread can probably figure out why I had to pour that idea down the drain.... Moral of this story: never, ever have dreams because they're always unatainable.

Wicked Pilot · Post by **Wicked Pilot** » 2003-01-24 11:49pm

Enlightenment wrote:Believe it or not at one point I was going to be an engineer. Anyone who's read this thread can probably figure out why I had to pour that idea down the drain.... Moral of this story: never, ever have dreams because they're always unatainable.

If you are under the impression that future engineers come out of their mother's womb doing calculus, you are wrong. Even the best and brightest of engineers had to start somewhere. Don't sell yourself short.

HemlockGrey · Post by **HemlockGrey** » 2003-01-25 12:00am

-reads Enlightment's post-

-puts the Superman cape back in the closet-

Enlightenment · Post by **Enlightenment** » 2003-01-25 09:56pm

Wicked Pilot wrote:Even the best and brightest of engineers had to start somewhere. Don't sell yourself short.

I'm 23 and I'm slashing and burning my way through my first year of university (college to US-types). I'm too old, it's too late, and I'm from the wrong background to start pursuing math & sciences now. I'm sticking with political science/international relations because that's an area where I can get marks in the 90-100 range rather than in the Ds.