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, we first calculate the distance traveled during the powered phase and then add the distance it coasts upward after the motor turns off. 1. During the powered phase (first 4 seconds), we use the equation $d = \frac{1}{2}at^2$, where $a = 20.0 \, m/s^2$ and $t = 4.0 \, s$. This gives us $d = \frac{1}{2} \times 20.0 \times (4.0)^2 = 160 \, m$.2. To find the velocity at the end of the powered phase, we use $v = at$, giving us $v = 20.0 \times 4.0 = 80.0 \, m/s$.3. For the coasting phase, we use the equation for motion under gravity $v^2 = u^2 + 2as$, where $v = 0$ (at the maximum height), $u = 80.0 \, m/s$, $a = -9.8 \, m/s^2$ (acceleration due to gravity acting downwards), and $s$ is the distance. Rearranging for $s$, we get $s = \frac{u^2}{2g} = \frac{(80.0)^2}{2 \times 9.8} \approx 327 \, m$.4. The total elevation is the sum of the distances in the powered and coasting phases: $160 \, m + 327 \, m = 487 \, m$.Therefore, the maximum elevation reached by the rocket is 487 meters.

A Test Rocket Is Fired Straight Up from Rest with a Net