Free fall: A test rocket is fired straight up from rest with a net acceleration of 20.0 m/s². After 4.00 seconds the motor turns off, but the rocket continues to coast upward with no appreciable air resistance. What maximum elevation does the rocket reach? A) 487 m B) 327 m C) 320 m D) 408 m E) 160 m

Question

Quizplus · Accepted Answer

To find the maximum elevation reached by the rocket, we first calculate the distance it travels while the motor is on and then the distance it travels while coasting upward until it comes to a stop.1. Distance traveled while the motor is on (using $d = \frac{1}{2}at^2$, where $a = 20.0 \, \text{m/s}^2$ and $t = 4.0 \, \text{s}$):$$d_1 = \frac{1}{2} \times 20.0 \, \text{m/s}^2 \times (4.0 \, \text{s})^2 = 160 \, \text{m}$$2. Velocity at the moment the motor turns off (using $v = at$, where $a = 20.0 \, \text{m/s}^2$ and $t = 4.0 \, \text{s}$):$$v = 20.0 \, \text{m/s}^2 \times 4.0 \, \text{s} = 80.0 \, \text{m/s}$$3. Distance traveled while coasting upward (using $d = \frac{v^2}{2g}$, where $v = 80.0 \, \text{m/s}$ and $g = 9.8 \, \text{m/s}^2$):$$d_2 = \frac{(80.0 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} = \frac{6400 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} = 326.53 \, \text{m}$$4. Total elevation reached is the sum of $d_1$ and $d_2$:$$d_{\text{total}} = 160 \, \text{m} + 326.53 \, \text{m} = 486.53 \, \text{m}$$Rounded to the nearest meter, the maximum elevation reached by the rocket is approximately 487 meters, which corresponds to choice A.

Free Fall: a Test Rocket Is Fired Straight Up from Rest