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a model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 75.7 m/s² for 1.61 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?

Explanation:

Step1: Calculate height during powered - flight

Use the equation $y_1 = v_0t+\frac{1}{2}a_1t^2$. Since $v_0 = 0$ (starts from rest), $a_1=75.7\ m/s^2$ and $t = 1.61\ s$.
$y_1=\frac{1}{2}\times75.7\times(1.61)^2$
$y_1=\frac{1}{2}\times75.7\times2.5921=97.93\ m$

Step2: Calculate velocity at the end of powered - flight

Use the equation $v = v_0 + a_1t$. Since $v_0 = 0$, $a_1 = 75.7\ m/s^2$ and $t = 1.61\ s$.
$v=0 + 75.7\times1.61=121.877\ m/s$

Step3: Calculate height during unpowered - flight

After fuel runs out, the rocket is in free - fall with $a=-g=- 9.8\ m/s^2$. At the maximum height, the final velocity $v_f = 0$. Use the equation $v_f^2 - v^2=2a_2y_2$.
$y_2=\frac{v_f^2 - v^2}{2a_2}=\frac{0-(121.877)^2}{2\times(-9.8)}=\frac{- 14853.07}{-19.6}=757.81\ m$

Step4: Calculate maximum altitude

The maximum altitude $y_{max}=y_1 + y_2$.
$y_{max}=97.93+757.81 = 855.74\ m$

Answer:

$855.74\ m$