Determine the velocity and the height of travel the
4 Mg rocket after a time of 4 s, starting from rest. A thrust force acting from the jets exerts a constant vertical force of 150 kN on the other end of the rocket. Neglect the loss of fuel during the lift.

1) 110.7 m/s & 0.22 km
2) 221.5 m/s & 0.32 km
3) 322.5 m/s & 0.45 km
4) 232.4 m/s & 1.25 km

The equation of motion can be written as
m (v_y)₁ + S F_y dt = m (v_y)₂
0 + 150 (10³) 4 - 4 (10³) (9.81) 4 = 4 (10³)v
v = 110.76 m/s = 110.7 m/s

So the velocity of the rocket after 4 s is v = 110.7 m/s

The equation of motion for the velocity can be
written as
v = v_o + a_ct
110.7 = 0 + a(4)
a = 27.69 m/s² = 27.7 m/s²

So the accleration of the rocket after a time of 4 s is a = 27.7 m/s²

The equation of motion for the distance travelled can be written as
s = s_o + v_ot + 1/2 a_ct²
s = 0 + 0 + 1/2 (27.7) (4)²
s = 221.5 m = 0.22 km

So the distance travelled by the rocket from the ground s = 0.22 km