PHYSICS HELP IN PROJECTILE!!!?

First find the velocity of the projectile at the top:

v = v0 cos(67) = 57.8 m/s

Use the conservation of momentum:

mv i = (m/3) v i - (m/3) v j + (m/3) v3

v3 = 2v i + v j

|v3| = sqrt((2v)^2 + v^2) = sqrt(5) v

= 2.24 * 57.8 m/s = 129 m/s

dir(v3) = atan(v/2v) = 26.6 degrees, above horizontal

To find the energy of the explosion, calculate the difference between the final and initial energies:

E = 1/2 (m/3) v^2 + 1/2 (m/3) v^2 + 1/2 (m/3) (sqrt(5) v)^2 - 1/2 mv^2

= (2/3) mv^2 = (2/3) (206 kg) (57.8 m/s)^2 = 459 kJ

1) it will have to equal the magnitude of the others in the other way so like

3 fragments

1 explodes -y direction

1 explodes x direction

so 4th quadrant

so the magnitude of the other fragment will be positive y, negative x direction

so it will be x^2 + y^2 =r^2

third fragment explodes at exactly 135 degrees (90+45, since sides are even)

and then the magnitude is 209.3036 m/s

A)209 m/s

B)135 degrees above horizontal... or 45 north of west

C)

1/2mv^2 is kinetics energy

so 2*K + 1times a different K

mv1^2 + 1/2mv2^2

206*148^2 + 1/2*206*209^2

K = 9,011,367 J

*notice the initial angle did not matter...

I've -re-read the question you badly worded it.

A & B conservation of momentum, all the velocities of the three parts have to add up to the velovity of the original projectile which is given by

Vh = V cos theta

C I'm assuming you mean the energy required to fire the projectile from a gun (or whatever).

E = 1/2 m . v^2 that's the easy part