Tackling the Difficult Death Problem in Physics Honors Class

[liljediboi]

hey. i am in a physics honors class, and we just had a test.

each test has 9 questions, 8 which are decent, and the last one is the "death problem" this problem is worth 20 point, double of the others (if you did the math, if you get die on the death problem, the highest score possible is 80).

anyways. the last death problem is as follows

a cannon fires a projectile (no air or wind) at 100 m/s at a 20 degrees angle north of west. it is fired down a "hill" (which is impossibly straight) the hill has a 20 degree angle north of east. find d (see picture.

talked to tons of people, nobody knows if they got it, and many did not even have time to do it. looks like a huge curve coming.

and the obvious question. WHAT IS THE ANSWER?? and please show work for full credit. haha

 

[HallsofIvy]

a cannon fires a projectile (no air or wind) at 100 m/s at a 20 degrees angle north of west. it is fired down a "hill" (which is impossibly straight) the hill has a 20 degree angle north of east. find d (see picture.

You do understand, don't you, that we CAN'T "see picture" because you didn't post it! I assume that "d" is the distance from the point at which the projectile is fired to where it hits the ground. The real point of the problem is to find where it hit.

You say "at 20 degrees angle north of west". That sounds like you mean it is aimed 20 degrees to the north of due west but if that is true, we don't know the upward angle at which the projectile is fired.

I'm going to take it that the cannon is aimed "north-west" (that is "45 degrees north of west", and that it is pointed 20 degrees above the horizontal. Also that the hill slopes downward in the same direction as the cannon is aimed (which makes the problem MUCH easier) with a 20 degree downward slope.

Here's how I would do the problem: We can ignore the direction. Since the cannon is fired in the same direction as the hill slopes, just do it as a standard two dimensional problem- take (0,0) as the position of the cannon and the x-axis in the direction the cannon is fired. First ignore the hill. Use the standard (parabolic) formula for a projectile launched at angle [theta] with initial speed v0:
x= v0 cos[theta] t, y= -(g/2) t²+ v0 sin[theta] t.
You can easily solve solve for t in terms of x and replace it in the formula for y to get y as a function of x. In a simple, "standard" problem, the projectile would hit where y= 0 so you would solve y= 0 for x.

Here, the ground slopes. Knowing that the ground slopes downward in the x direction, find the equation of the straight line:
it is y= (tan[theta])x where [theta] is the angle the line makes with the x-axis (in this problem it is -20 degrees).

The projectile "hits" the ground when it crosses that line.
Solve the two equations y= f(x) for the projectiles motion and
y= mx for the ground simultaneously to find the point at which the projectile hits the ground. Once you know the coordinates of that point, you can find d, the distance between (0,0) and that point.

 

[liljediboi]

attachment is working now. it is not very accurate, as my friend has pointed out, but give me a break, i used mspaint.

anyways the cannon fires 100 m/s [W 20 N]

 

[HallsofIvy]

Then you have not given us the angle "of elevation" of the cannon barrel and it is impossible to answer this.

 

[liljediboi]

whatever. found out how to do it, insanely easy, its 1440 m

 

[HallsofIvy]

"whatever".

Ah, the enthusiasm!

I glad it was "insanely easy". You do understand, don't you that you never posed a solvable problem here?

Assuming that the problem was as I interpreted it, 1440 m is close.
I get 1486 meters.