What Angle and Velocity Are Needed for a Cannon to Hit a Falling Target?

[dekoi]

A small cannon (angle manipulative) produces an unknown force on a bullet. Simultaneously, a plate is dropped from h`1 . The cannon is d`x 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

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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. I am in deep fatigue and can't think correctly. Anyone willing to help out?
Thank you.

 

[dekoi]

bump!

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

 

[dekoi]

Does anyone have any advice?

 

[marlon]

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

 

[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

 

[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