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