- #1

ph123

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I know this has something to do with p=m(v_center of mass), but I'm not really sure how to deterine the acceleration of the center of mass. It is constant up until the explosion, but then the explosion stops one half of the project dead in its tracks where it falls straight down, whereas the other half of the projectile is accelerated with an equal but opposite direction as the first half of the projectile. I do not know how to quantify that.

I was able to get answer with the following equations, though I am unsure as to whether I used a correct approach.

The equation for the max height reached by a projectile is :

m.h. = [V_0^2(sin^2(theta)]/2g

m.h. = [(84.0 m/s)^2(sin^2(65))]/19.60 m/s^2

m.h. = 295.7 m

I deduced this distance along the ground using (R = the hypotenuse):

Rsin(theta) = height

R = height/sin(theta)

R = 295.7m/sin(65)

R = 326.27m

Rcos(theta) = distance along ground

(295.7m)cos(65)

=137.89m

This is the point to which I calculated the half of the projectile that falls vertically down from the max height to land. To find the position that the center of mass lies along the ground, I used the equation for max range of a projectile.

m.r. = [(V_0)^2(sin(theta))]/g

m.r. = [(84.0 m/s)^2(sin(65))]/9.8 m/s^2

m.r. = 652.54m

This is where I calculated the center of mass of the two halves of the projectile to be. To get the distance of the other half of the particle (that was accelerated forward), I subtracted the distance of the first half by the position of the center of mass, then added this value to the center of mass.

652.54m - 137.89m = 514.65m

514.65m + 652.54m = 1167.19

The distance the particles lie from each other is the difference of the two distances:

1167.19m - 137.89m = 1029.3m

Is this a correct approach? If not, then how do I figure out the acceleration after the explosion?