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you launch a model rocket from ground level. it moves directly upward with a constant acceleration of $67.0\\ \text{m/s}^2$ for 1.25 seconds, at which point it runs out of fuel. assuming air resistance on the rocket is negligible, what is the maximum altitude (above the ground) achieved by the rocket?
enter a number: ______ m
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Explanation:

Step1: Find velocity at fuel depletion

$v = v_0 + a t$
$v = 0 + 67.0 \times 1.25 = 83.75 \, \text{m/s}$

Step2: Find height during acceleration

$h_1 = v_0 t + \frac{1}{2} a t^2$
$h_1 = 0 + \frac{1}{2} \times 67.0 \times (1.25)^2 = 52.34375 \, \text{m}$

Step3: Find height after fuel runs out

At max height, final velocity $v_f=0$. Use $v_f^2 = v^2 - 2 g h_2$
$0 = (83.75)^2 - 2 \times 9.81 \times h_2$
$h_2 = \frac{(83.75)^2}{2 \times 9.81} = \frac{7014.0625}{19.62} \approx 357.5 \, \text{m}$

Step4: Sum heights for max altitude

$H = h_1 + h_2$
$H = 52.34375 + 357.5 \approx 409.84 \, \text{m}$

Answer:

$\approx 410 \, \text{m}$ (or $409.8 \, \text{m}$ for more precision)