Self-education is, I firmly believe, the only kind of education there is. -Isaac Asimov

Trajectory, projectile motion

Posted: November 5th, 2011 | Author: | Filed under: Algebra, Mathematics, Physics | Tags: , , , | No Comments »

One of the most annoying problems I’ve had is one of Trajectory and Projectile motion. The reason that these problems are so frustrating though is not because of the actual physics (which conceptually is quite easy) but because of the algebra. It can be quite confusing knowing where to move things around. In projectile motion problems, this becomes even harder because there are some intimidating trig functions that need to be resolved.

So let’s do one as an example which is a bit intimidating just because of the algebra, I’ll do the step by step derivation.

During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. Figure 4-51 shows a cross section of Mt Fuji in Japan.

(a) At what initial speed would a bomb have to be ejected, at angle theta_0=35^circ to the horizontal, from the vent at A in order to fall to the foot of the volcano at B, at vertical distance h = 3.30 km and horizontal distance d = 9.40km? Ignore for the moment , the effects of air on the bomb’s travel. (b) what would be the time of flight?

volcanoes are scary.Luckily, they’ve given us a picture to see what is happening.

 

 

 

First let’s take a look at what the question is looking for and then we’ll delve into how to work it. The question specifically is asking for the speed as it leaves the volcano.  So what we are looking for is the MAGNITUDE of the initial velocity. But this is kind of perplexing because we’re not given a lot of information, we’re given the height (h) the distance (d) and the angle. Let’s make a table of what we have and what we need.

[table id=2 /]

Not a lot of information, but we have an equation to help us with this. The path of the trajectory of a projectile is given by the equation:

y=tan(theta)x-frac{gx^2}{2(V_0cos(theta)^2)}

We know that we are solving for the magnitude of the initial velocity. so let’s do that

y=tan(theta)x-frac{gx^2}{2(V_0cos(theta)^2)}

 

tan(theta)x-y=-frac{gx^2}{2(V_0cos(theta))^2}

 

tan(theta)x-y=-frac{gx^2}{2(V_0cos(theta))^2}

 

2(V_0cos(theta))^2tan(theta)x-y=-gx^2

 

V_0cos(theta))^2=-frac{gx^2}{2(tan(theta)x-y)}

 

V_0^2=frac{x^2}{cos(theta)^2}frac{g}{ 2(tan(theta)x-y)}

 

V_0=frac{x}{cos(theta)}sqrtfrac{g}{ 2(tan(theta)x-y)}

Note how the square roots were saved for last in the equation, this is because these were the trickiest parts.

But look now! We are 90% of the way done with part a! Now it’s all just plugging in our known values!

V_0=frac{9400m}{cos(35^circ)}sqrtfrac{-9.8frac{m}{s^2}}{ 2(tan(35^circ)9400m-(-3300m))}

This comes out to about 255frac{m}{s}

Easy! See that we just needed to think about the algebra, and the rest of the problem came together!

Now lets do part b. In this I’ll just give you the equation since it’s a simple manipulation

x=V_0cos(theta)t

 

frac{x}{V_0cos(theta)}=t

Hey we have all these values!

frac{9400m}{255frac{m}{s}cos(35^circ)}= 45s

It’s solved! Go get yourself a drink!

Problem is from 91 in Halliday & Resnick’s Fundamentals of Physics