# Homework Help: An interesting Rifleman Problem (Projectile Motion)

1. Sep 15, 2006

### Tony Zalles

Hi,

So this is the problem out of Fundamentals of Physics by Haliday Resnick and Walker (Seventh Edition.

By the way this problem would all under a first year calculus based mechanics course.

I know calculus through multivariable...but thats not needed for this problem...at least I don't think so.

In any case here it is:

31. A rifle that shoots bullets at 460 m/s is to be aimed at target 45.7 m away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that this bullet hits dead center.

Ok... So I know that there is actually a derivation known as the "Rifleman's Rule" for projectile motion (I believe) can be applied to this problem....but it is a very tedious derivation...whose end result is an approximation only (which works out fine because it is intended to be an approximation due to the very small angles that would be used) but that aside, I believe there is a simpler way to work out this problem. But I fail to see how...

Ok here is what I did...

Ok the barrel has to be fired at some [initial angle] and its path is parabolic....
this gives me some useful insight, the time for bullet to go half the horizontal distance if doubled gives the full horizontal distance...and that the [initial angle] when measured from the +x is related to the final angle by taking

[180 degrees - [inital angle]] = [final angle]
(taking final angle to be measured from +x aswell.

Ok so listing the magnitudes:

[inital velocity] = 460
[final velocity] = ?

---------------------------------

[inital x position] = 0
[inital y position] = 0
[final x position] = 45.7
[inital angle] = ?
[final angle] = ?

[initial velocity in x] = ?
[final velocity in y] = ?

[final velocity in x] = [initial velocity in x] = ?
[final velocity in y] = ?

Ok velocity in x is constant
Velcity in y follows a free fall system

I already derived the initial components of velocity....

cos [inital angle] * Vo = x component of initial velocity
sin [inital angle] * Vo = y component of initial velocity

I set up two position functions each for x and y plugged in the above derivations.

x(t) = 0 + Vx0 * [delta t] + 0
y(t) = 0 + Vy0* [delta t] - (1/2)*g*{[delta t]^2}

I also then set up two equations from the form

Final velocity = initial velocity + at

So two equations of the above form each for x and y....and you know by I figured I could see something that if I solved for in of these equations plugged back in to another....something would cancel out and I would actually be able to solve for at least one of the variable....

or be able to set up a system of equations for my variables....however....I'm not be able to solve this using either approach....

so...yea. any help would be nice :)

Thanks,

-Tony Zalles

2. Sep 15, 2006

### HallsofIvy

You have
x(t) = 0 + Vx0 * [delta t] + 0
y(t) = 0 + Vy0* [delta t] - (1/2)*g*{[delta t]^2}

You know that the target is 45.7 m away so you can solve
45.= Vx0 * [delta t] for delta t as a function of Vx0.
You can then put that into 0 = Vy0* [delta t] - (1/2)*g*{[delta t]^2} and get an equation relating Vx0 and Vy0. From that you should be able to calculate the initial angle. Now use tangent of that angle to find how high above the target you must aim.

3. Sep 15, 2006

### Tony Zalles

ok,...

So for the initial theta I end with the following:

2*(sin(inital angle))*(cos(initial angle)) = [(final x)(g)]/(initial velocity)

And with figures:

2*(sin(inital angle))*(cos(initial angle)) = [(45.7)(9.8)]/(460)

Which reduces to...

sin(2*(initial angle)) = [(final x)(g)]/(initial velocity)

solving for (inital angle)

gives,

(initial angle) = 38.4 degrees

Now, I believe I can set...

tan(initial angle) = (final y)/(final x)

However in deriving this relationship I already assumed that the final y position was zero in order to be hit dead on to the target...

so...this problem is annoying they didn't even give a figure/picture...

Although I really appreciate the help HallsofIvy :)

The answer in the back of the book is 4.84 cm.

And I'm not getting that....and I know the angle is right because if I use it to calculate the Vx0 and Vyo and take the square of each and then the square root of their sum I end up back with 460....so yea...still a little stuck....

Again any help would be most appreciated :)

-Tony Zalles

4. Sep 15, 2006

### andrevdh

The range of a projectile can be shown to be given by

$$R = \frac{V^2\sin(2\alpha)}{g}$$

where V is the launch velocity and alpha the initial launching angle.

5. Sep 15, 2006

### Tony Zalles

Never mind,,....numerical error on my part....it all works and I get the answer in the back of the book.

Thanks.

-Tony Zalles