Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.2 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.80 m/s2 until it reaches an altitude of 1190 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)

Answer 1

Task:

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of $80.2\mathrm{m / s}$ at ground level. The engines then fire, and the rocket accelerates upward at 3.80 m/s2 until it reaches an altitude of $1190\mathrm{m}$ . At that point its engines fail, and the rocket goes into free fall, with an acceleration of $-9.80\mathrm{m / s2}$ . (You will need to consider the motion while the engine is operating and the free-fall motion separately.)

Solution:

$$s _ {0} = 0, v _ {0} = 8 0. 2 \frac {m}{s}, a = 3. 8 \frac {m}{s ^ {2}}, s = s _ {0} + v _ {0} t + \frac {a t ^ {2}}{2}$$ $$s _ {1} = 0 + 8 0. 2 \frac {m}{s} \cdot t + \frac {3 . 8 \frac {m}{s ^ {2}} \cdot t ^ {2}}{2}, 1 1 9 0 m = 0 + 8 0. 2 \frac {m}{s} \cdot t + \frac {3 . 8 \frac {m}{s ^ {2}} \cdot t ^ {2}}{2}$$

$t = 11.6323s$ - the time after launch with working engines

$$v _ {1} (t) = v _ {2 0} = v _ {0} + a t = 8 0. 2 \frac {m}{s} + 3. 8 \frac {m}{s ^ {2}} \cdot 1 1. 6 3 2 3 s = 1 2 4. 4 0 2 7 \frac {m}{s}$$ $$s _ {2} = 1 1 9 0 m + 1 2 4. 4 0 2 7 \frac {m}{s} \cdot t - \frac {9 . 8 \frac {m}{s ^ {2}} \cdot t ^ {2}}{2}$$ $$0 = 1 1 9 0 m + 1 2 4. 4 0 2 7 \frac {m}{s} \cdot t - \frac {9 . 8 \frac {m}{s ^ {2}} \cdot t ^ {2}}{2}$$

$t = 32.7939$ s - the time after engines fail till the crash

$t = 44.4262s$ -total time of the motion

Mathematical model of the motion: