# Projectile Motion Lab Question

1. Sep 23, 2004

### dekoi

A small cannon (angle manipulative) produces an unknown force on a bullet. Simultaneously, a plate is dropped from h1 . The cannon is dx away from the target's horizontal location. What angle is needed, as well as the minum velocity, for the bullet to hit the falling target before its impact with the floor h`1 below. this is similar to this animation: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/mz.gif

----

I am told that i have to do this "algebraically", using ONLY the 5 kinematic equations (and their projectile equivalent) in order to prove the launch angle. You can only use dx, dy, and h in these equations (and i am guessing variables such as g are also allowed).

I am not sure what is meant by algebraically (specifically); although my method seems awkwardly long, yet i am guessing it would work.

firstly, i calculated the time needed for the object falling vertically to fall (square root: 2h/g = t). I used that equation along with vx = dx / t, to figure out the minmum horizontal velocity. v2x = v1x.
dy = v1y(t) + 1/2gt^2 .

Method 1:
0 (d2y-d1y = 0) = v1xtanX(t) + 1/2gt^2
Rearrange to solve for angle X.

Method 2:
0 = v1tsinX + 1/2gt^2
Rearrange for X.

Im not really sure about this. Im in deep fatigue and cant think correctly. Anyone willing to help out?
Thank you.

2. Sep 23, 2004

### dekoi

bump!

I'm sorry, but i haven't heard the policy against or for 'bumping'. Would anyone mind informing?

3. Sep 24, 2004

### dekoi

4. Sep 24, 2004

### marlon

what is the distance between the canon and the target ??
is it dx ???
marlon

5. Sep 24, 2004

### marlon

You are right on the x-component of the velocity. It is dx/t and t = sqrt(2h/g).

But are you sure the difference between target and source is dx???
marlon

6. Sep 24, 2004

### marlon

For the Y-component just write y as a function of x. Just substitute the t-variable in the equation for y by t = x/v_h with v_h the horizontal velocity of the bullet at t = 0. Then calculate the derivative of y to x and after you substituted the distance between source-target just solve this equation for the y-component of the velocity...

The angle is tan(X)=(v_y/v_x) y-component devided by x-component of the velocity.

This is just a suggestion. There are other ways to solve this question. I suggest you also try the other ones...for eeuuhh fun...

marlon