Test your Knowledge

[sjohnson]

Little game to test knowledge of projectile..

http://www.globalspec.com/trebuchet...=8E4795&[email protected]

 

[Peter Nachtwey]

Little game to test knowledge of projectile..

http://www.globalspec.com/trebuchet...=8E4795&[email protected]

Good game. It is a little simplistic because it doesn't take into account the rotational inertia of the swing arm.
I would generate a work sheet to do the calculations before playing.

 

[daba]

I would generate a work sheet to do the calculations before playing.

And I would just play the game, for fun !

Peter, your worksheet would probably give more realistic results than the math in the game , lol : you'd end up with a perfect theoretical solution that doesn't work so well. Perhaps you'd get your fun doing the maths, and the game after would be inconsequential ...

 

[Peter Nachtwey]

I would take into account wind resistance too if it were a realistinc problem

I would have to think about how to take the rotation of the earth into account.

You start with m*g*h=m*v^2/2 but in reality one would have to use
m*g*h=I*omega^2/2. Now one needs to know the length of the throw are to find the relationship between omega and velocity.
Then you need to big an angle, call it alpha. The distance over ground is v*cos(alpha)*deltat. Now deltat needs to be computed. That is simple. The vertical component of v is v*sin(alpha). While traveling up the v*sin(alpha) is opposed by gravity. So v*sin(alpha)=g*timeup. deltat=timeup+timedown. So deltat=v*sin(alpha)*2/g.

Here is a another detail that is over looked. If the projectile is thrown at 45 degrees then the throw would be from sin(45 deg)*length of the swing arm behind the axis of rotation. So where is the distance measured from, the axis of rotation or where the projectile leaves the throw arm?

There are too many 'holes' in this problem to take serious. I would have to play 20 questions and it isn't worth my time.

 

[daba]

Here is a another detail that is over looked. If the projectile is thrown at 45 degrees then the throw would be from sin(45 deg)*length of the swing arm behind the axis of rotation. So where is the distance measured from, the axis of rotation or where the projectile leaves the throw arm?

It would have to be a constant for different launch angles, so would have to be the ground position of the swing-arm axle. Pretty much like the bowlers popping crease in cricket - the bowler may well release the ball in front or behind this line.