The rocket attains a speed of 60 m/s in a matter of 6.3 s, so it is undergoing an acceleration a of
a = (60 m/s - 0) / (6.3 s) ≈ 9.52 m/s²
During launch, the only forces acting on the rocket are directed vertically:
• its weight (pointing down) with magnitude W,
• drag (pointing down) with magnitude 50 N, and
• thrust (pointing up) with magnitude T.
(a) By Newton's second law, the magnitude of the net force is proportional to the acceleration according to
∑ F = m a
where m is the mass of the rocket, so the magnitude of the net force is
(8 kg) (9.52 m/s²) ≈ 76.2 N
and it is directed upward.
(b) Expanding the equation from part (a) gives
T - W - 50 N = (8 kg) (9.52 m/s²)
The weight is
W = (8 kg) g = (8 kg) (9.80 m/s²) = 78.4 N
so the thrust force is
T ≈ 76.2 N + 78.4 N + 50 N ≈ 205 N
(c) We found this earlier, a ≈ 9.52 m/s².
(d) The rocket's height y at time t is given by
y = 1/2 a t ²
so that after 6.3 s, it will have reached a height of
y = 1/2 (9.52 m/s²) (6.3 s)² = 189 m
