# Velocities and momentum after explosions

#### Delta Sheets

1. The problem statement, all variables and given/known data
After several large firecrackers have been inserted into its holes, a bowling ball is projected into the air using a homemade launcher and explodes in midair. During the launch, the 5.43-kg ball is shot into the air with an initial speed of 12.9 m/s at a 39.9° angle; it explodes at a peak of its trajectory, breaking into three pieces of equal mass. One piece travels straight up with a speed of 3.83 m/s. Another piece travels straight back with a speed of 1.71 m/s.

a) What is the velocity of the third piece?
Answer calculated: 31.63 m/s

b) What is the direction of the velocity of the third piece?

2. Relevant equations
Momentum is conserved
m1v1 + m2v2 + m3v3(initial)=m1v1 + m2v2 + m3v3(final)

3. The attempt at a solution
I figured since there is no momentum in the y direction at the time of explosion, the y momentum going up must equal the y momentum going down, which is 6.93 kg*m/s.
With this is took the momentum of the third piece (31.63)(1.81)= 57.25 kg*m/s, then used the tangent inverse tan^-1(-6.93/57.25) to find the angle, but it did not give me the right answer.

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