posted by Anonymous .
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 5.33-kg ball is shot into the air with an initial speed of 11.3 m/s at a 42.9° 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 3.45 m/s. Another piece travels straight back with a speed of 2.39 m/s. What is the velocity of the third piece?
A = 3.45 m/s[90o]
B = 2.39 m/s[270o]
C = V m/s
3.45[90o]+2.39+V = 0 @ the peak of
3.45*cos90+2.39*cos270+i3.45*sin90+i2.39*sin270 + V = 0
0+0 + 3.45i-2.39i + V = 0
0 + 1.06i + V = 0
V=-1.06i = 1.06 m/s[270o]=1.06m/s[-90o].