Topic: Engineering Mechanics
A rocket is launched vertically up. It attains the velocity of $200~m/s$ at an altitude of 1.125 kms. At this point, its booster stage is separated without any relative velocity. Find the maximum altitude the booster stage attains and the time duration between separation and its impact on ground. Consider no drag effect on the separated booster stage and $g=10~m/s^{2}$.
- A. $Y_{max}=1.125~km$ and $t=20s$
- B. $Y_{max}=3.125~km$ and $t=25s$
- C. $Y_{max}=2.0~km$ and $t=45s$
- D. $Y_{max}=3.125~km$ and $t=45s$
Answer: D. $Y_{max}=3.125~km$ and $t=45s$
Explanation: Initial upward velocity of separated booster, $u = 200~m/s$.
Altitude at separation, $h_0 = 1125~m$.
Maximum additional height reached from separation point = $\frac{u^2}{2g} = \frac{200^2}{2 \times 10} = \frac{40000}{20} = 2000~m = 2.0~km$.
Total maximum altitude $Y_{max} = 1.125~km + 2.0~km = 3.125~km$.
Time taken to reach maximum altitude from separation, $t_1 = \frac{u}{g} = \frac{200}{10} = 20~s$.
Time taken to fall freely from $Y_{max}$ to ground, $t_2$: $Y_{max} = \frac{1}{2}gt_2^2 \Rightarrow 3125 = 5t_2^2 \Rightarrow t_2 = \sqrt{625} = 25~s$.
Total time duration $t = t_1 + t_2 = 20 + 25 = 45~s$.