Question

A 2.2-kg model rocket is launched vertically and reaches an altitude of 70 m with a speed of 30 m/s at the end of powered flight, time t = 0. As the rocket approaches its maximum altitude it explodes into two parts of masses mA = 0.77 kg and mB = 1.43 kg. Part A is observed to strike the ground 80 m west of the launch point at t = 6 s. Determine the position of part B at that time.

Answer 1

Answer:

$r_b= (30.8\hat{i} + 69.96 \hat{j}) m$

Explanation:

given,

mass = 2.2 kg

altitude(r₀) = (70 j) m

speed = 30 m/s

m_a = 0.77 kg

m_b =1.43 kg

part A strike ground (r_a)= (80 i) m

t = 6 s

$r = r_0 + v_ot-\dfrac{1}{2}gt^2$

$r = 60\hat{j} + (30\hat{j})\times 6-\dfrac{1}{2}\times 9.8 \times 6^2$

r = 63.6 j m

by conservation of energy

$mr = m_ar_a+m_br_b$

$2.2\times 63.6\hat{j} = 0.77\times (-80 \hat{i})+2\times r_b$

$r_b= (30.8\hat{i} + 69.96 \hat{j}) m$