Question: A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of $76.4\ \text{m/s}^2$ for $1.63$ seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?
1. **State the problem:**
A rocket accelerates upward from rest with acceleration $a=76.4\ \text{m/s}^2$ for $t_1=1.63\ \text{s}$, then coasts upward until it reaches maximum altitude. We want to find the maximum altitude above the ground.
2. **Identify known values:**
- Initial velocity $u=0$ (starts from rest)
- Acceleration during powered flight $a=76.4\ \text{m/s}^2$
- Time of powered flight $t_1=1.63\ \text{s}$
- Acceleration after fuel runs out $=0$ (only gravity acts, but air resistance ignored, so velocity decreases due to gravity $g=9.8\ \text{m/s}^2$ downward)
3. **Calculate velocity at end of powered flight:**
$$v = u + at_1 = 0 + 76.4 \times 1.63 = 124.532\ \text{m/s}$$
4. **Calculate altitude reached at end of powered flight:**
$$s_1 = ut_1 + \frac{1}{2} a t_1^2 = 0 + \frac{1}{2} \times 76.4 \times (1.63)^2 = 0.5 \times 76.4 \times 2.6569 = 101.4\ \text{m}$$
5. **Calculate additional altitude during coasting phase:**
After fuel runs out, the rocket moves upward with initial velocity $v=124.532\ \text{m/s}$ and decelerates due to gravity $g=9.8\ \text{m/s}^2$ until velocity becomes zero at maximum altitude.
Use formula:
$$v^2 = u^2 + 2as$$
Here, final velocity $v=0$, initial velocity $u=124.532$, acceleration $a=-9.8$ (negative because gravity opposes motion), solve for $s_2$:
$$0 = (124.532)^2 + 2 \times (-9.8) \times s_2$$
$$2 \times 9.8 \times s_2 = (124.532)^2$$
$$s_2 = \frac{(124.532)^2}{2 \times 9.8} = \frac{15510.9}{19.6} = 791.1\ \text{m}$$
6. **Calculate total maximum altitude:**
$$s_{max} = s_1 + s_2 = 101.4 + 791.1 = 892.5\ \text{m}$$
**Final answer:** The rocket reaches a maximum altitude of approximately $892.5\ \text{m}$ above the ground.