# Attn. Physics People

In my engineering physics class, This was one of the problems in the book.

A_______/__
|…/
|…/
|…/
|…/
|…/
|X./
|/
/B_________

Object A is traveling at 3.0m/s along the horizontal line. Object B starts at the instant that A crosses the vertical line with an acceleration .40m/s2. The two horizontal lines are 30m apart. What is angle X so the two object will collide?

I started off by setting up some equations:
(3.0)t = (1/2)(.40sinX)(t^2) setting the horizontal distance equal
tan(X) = 3t/30 To find angle

I solved the first equation for t = 15/(sinX) Then I put that into the other equation to get:
tan(X) = 3/(2sinX)

I put each side into my calculator and found that the intersect was 60 degrees, the correct answer.

So my question is: What is the easier way do solve this? (there has to be one) Or, how do you solve the last equation.

The dumb part is that it wasn’t even assigned, I was going through the book and now I must figure it out!

seems like a good way to solve it to me.
There’s no way to solve it without two equations since you have two unknowns (t and X).

this is soooooo easy!

You have a rectangular triangle. the vertical side is 30
the horizontal (upper) side is = 3*t
the hypothenuse = (1/2)0.4t^2

so we have (30)^2 + (3t)^2 = (0.2*t^2)^2

resolve that and you get t^2 = 300, t=17.32

so the upper side is 51.96 and the hypothenuse 60

the angle is arctan (51.96/30) = 60 degrees F (not hot but not cold either)

No, in fact that is one of the smarter things you can do - find a problem that isn’t assigned, think about it, and work it out. It shows great promise about your intellect, IMO.

sailor-
I had convinced myself that it couldn’t be that easy! I was digging through my calculus book looking for a trig identity to solve this. BTW, the instructor got me started off the way I showed.

Anthracite-
Thanks for the compliment.

I am always surprised at how people tend to look for complicated solutions when there are simpler ones. I have taught celestial navigation, which requires using tables, graphics etc, It is not excessively complicated and people can learn it in a few hours although it does require some practice to set everything in.

Anyway, I have tried this several times. I ask these people a very simple problem and inevitably they go off in search of the solution in the most complicated directions when the answer is so simple. But they have just learnt a ton of stuff and they feel their newly acquired knowledge must be put to good use.

Here is the problem:

My boat can make 10 knots through the water. The current set and drift are 030T and 2.5 knots. My desired true course to my next waypoint is 075T. What course should I steer in order to offset the current and travel the desired course?

Just to clarify: Courses are reckoned from true north clockwise so, for instance, 030T means the current flows to roughly the NE.

This is so simple to resolve! And yet I see people who have resolved much more complicated stuff stumped by it.

Bonus question. My next waypoint is 20 NM away (bearing 075T as has been said above). How long will it take me to get there?

Since we’re going to Cuba anyway…

The drawing in the OP reminded me of a lie my Physics teacher told me in high school. He said that if I had a slingshot, and I aimed it at a monkey sitting in a tree (never mind why), and I shot at it, and at the instant I released the projectile the monkey had a heart attack and fell off the branch, I would hit it on the way down since both the monkey and my pebble are falling at the same rate (namely, 9.8m/s/s).

This sounds like a crock. He spent a lot of time trying to get me to believe it.

He failed.

sdimbert, for once your teacher was right (I know after the many instances of teachers being wrong we’ve been having, this is hard to believe). You can just take my word for it or someone will be with you shortly to give you a detailed explanation.

All projectiles drop in flight. With a handgun at short range, this drop is tiny and you can aim the thing like a laser beam. With a rifle at long ranges, this drop is significant and you have to adjust the sights to compensate, i.e. you have to AIM HIGH.

Now, with a slingshot at long range, we can assume quite a large projectile drop, say a yard.

If you aim straight at the monkey, with the aid of your laser sight, you will actually miss him, the shot will pass a yard below. To hit him, you’d have to aim a yard high.

However, if you aim straight at the monkey and he has a heart attack and drops out of the tree, then you will hit him because he drops a yard, just as your projectile drops a yard.

What your physics teacher didn’t tell you was that if you aimed straight at the monkey and he stayed put, you’d miss. You could have also argued that you have a kickass slingshot with only a couple of inches of drop so you’d nail the monkey whatever, or brought up air resistance to really spoil his day!

A similar problem (even if id don’t look simular to you)

A bomber airplane wants to hit a target A. It flies in a straight line that will take it directly over the target. It (obviously) needs to release the bombs before it overflows the target and the bomb will hit the target at exactly the moment when the airplan is overhead. It does not matter how high or low the airplane was flying. (we are neglecting air resistance, of course, we do not want to spoil the fun)

So, I am the pilot. i am flying at altitude X feet. I calculate how long it would take for an object dropped from an altitude of X feet to hit the ground below and find out it is t seconds. I release my bomb t seconds before I overfly the target and the bomb should hit it.

now, who is going to tackle my navigation problem? Class? Anyone?

sailor

x = x + vt + (1/2)gt^2

no initial velocity or position, so

x = (1/2)gt^2

solve for t: t = (2x/g)^(1/2) I don’t know how to show the square root any other way.

What do the bearings translate to in degrees?

I am making a few assumptions:
1-a knot is a nautical mile per hour
2-030T means 30 degrees off of north

So: I plotted the current and the direction of the intended bearing on a coordinate plane. The direction of the current was 45 degrees up from the intended bearing.

Then I found the vertical component of the current by 2.5sin45 = 1.77 knots. To go along the inteded path the boat’s vertical velocity must cancell that out, so: its vertical component is -1.77 knots.

The components of the boat form a right triangle.
One side is -1.77, the hypotonuse 10, and the other side is the one along the intended path. It is 9.84 knots (pythagorean theorem)
The angle is tan(x) = 1.77/9.84 x = (about) 10 degrees

10 degrees + 45 + 30 = 85 degrees or 085T

Bonus: time to go 20 NM velocity along the intended path is 9.84 + 1.77 = 11.61 knots

So t = d/v = 20/11.61 = 1.72 hours.

How’d I do?

Yes, your soloution is the easy eligant one. But “rectangular triangle”? A four-sided three-sided object? I think you meant right triangle.

In Human Action Von Mises refers to it as a rectangular triangle as well, and Von Mises seems pretty smart (as does sailor), so I’m assuming that it’s a correct term, even if uncommon.

matt, sailor, et al,

Thanks for the explanations. I knew that if someone took the time to use little words I would eventually understand it!

The point of comprehension for me came with the idea that if I aimed right at Curious George, I would miss. Thank you.

So, my mind wanders…

Say I had a high-powered rifle aimed along a line parallel to the ground, but six feet above it. I place a single bullet in the chamber and another one on top of the end of the barrel so that, as the first exits the barrel, it will dislodge the second and set it falling to the floor.

Assuming that:
[ul]
[li]the first bullet leaves the gun at exactly the same time that the second falls,[/li][li]I am in a vacuum, and[/li][li]the bullets are of exactly equal mass,[/li][/ul]
They will both hit the ground at the same time, won’t they?

sdimbert, you also need to assume that the surface is level and that the gravitational field is uniform, but other than that, you’ve got it.

AWB, Rectangular means in general anything with a right angle. Rect means right, angular, angle. Rectangular triangle is a triangle with a 90 degree angle.

In common English usage you were thinking of rectangle but even this is imprecise and the strictly correct way of expressing what you have in mind would be “rectangular parallelepiped”.

hightechburrito:

>> I am making a few assumptions: 1-a knot is a nautical mile per hour

Well, yes, any other assumption would be wrong

>> 030T means 30 degrees off of north

Well, this was clarified in my original question when I said “Courses are reckoned from true north clockwise so, for instance, 030T means the current flows to roughly the NE”. Yes, in nautical terms all courses and directions are reckoned clockwise from North. If true north 030T, if magnetic north 030M, if compass north 030C etc. To keep the problem simple we can just assume all bearings are true.

Now the solution. This is best done graphically. Because I cannot post a graphic here, I have posted the solution here.
Sdimbert, your bullets will hit the ground at the same time even if they are of different mass. Only the other two conditions are necessary.

PS. man is the board slow today!

Actually, sdimbert, your teacher was wrong, and Cecil explains why.

http://www.straightdope.com/classics/a5_201a.html

From the right triangle: sin[sup]2/sup=1-cos[sup]2/sup=225/t[sup]2[/sup] and cos(x)=150/t[sup]2[/sup]

So, 0=1-1.5*cos(x)-cos[sup]2/sup=(0.5-cos(x)) * (2+cos(x))

cos(x)=0.5: x=60 degrees or cos(x)=-2: x= pi-i*log(2-sqrt(3))

Manlob, can you do that step by step? I do not follow it but, in any case, I do not believe it is simpler or more direct than what I did.