Answer:
speed is -20.0155 m/s
and direction is downward
Explanation:
given data
mass of rocket = 1500 kg
rocket accelerates upward = 10 m/s²
time t = 2 s
maximum height = 530 m
to find out
speed and direction of the heavier fragment just after the explosion
solution
we use here principle of conservation of momentum that is
Mt Vt=M1 V1 + M2 V2 ................1
so here
first we find the initial velocity that is express as
v = a × t .....................2
v = 10 × 2
v = 20 m/s
and distance S
S =u × t + 0.5 × a × t² ...................3
S = 0 + 0.5 × 10 × 2²
S = 20 m
and now we find speed of the lighter piece that is
v² = v1² + 2 × a × S .................4
0 = v1² + 2 × -9.81 × (530-20)
v1² = 10006.2
v1 = 100.031 m/s
so now put all these in equation 1 we get
Mt Vt = M1 V1 + M2 V2
1500 × 20 = (1500 × 1/3 ) × 100.031 + (1500 × 2/3) + V2
30000 = 50015.5 + 1000 V2
V2 = -20.0155 m/s
so speed is -20.0155 m/s
that mean vertical component of second fragment will be move downward