<p>A rocket accelerates vertically upward at 14m/s^2 for 30s then the engine goes off. What is the maximum height reached by the rocket?</p>
<p>Here’s a way to get started: draw a velocity vs time graph for the situation. You will see that it has two sections, both linear. First it slopes upward from zero with a slope of 14, fo 30 seconds. Then it slopes downward back to zero at a slope of -10 (or -9.8 if you prefer.) Use the graph and the slopes to figure out whatever else you need to find the area under the graph. (I like this method bc you don’t need any kinematics equations.)</p>
<p>gizmodo’s answer</p>
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<p>The height reached before engine goes off is
h<em>1 = a t</em>1^2/2, where a = 14 m/s^2, t<em>1 = 30 s.
In the next part of motion the rocket has initial velocity v</em>1 = a t<em>2 and experiences gravitational acceleration g = 9.81 m/s^2.
The maximum height is reached when velocity becomes zero at time t</em>2 = v<em>1/g (because v(max height) = 0 = v</em>1 – g t<em>2). The distance travelled in this part of motion equals h</em>2 = v<em>1 t</em>2 - g t<em>2^2/2 = v</em>1^2/2g = a^2 t_1^2/2g.</p>
<p>So, the maximum height reached by the rocket is
H<em>max = h</em>1 + h<em>2 = a t</em>1^2/2 + a^2 t<em>1^2/2g = (1 + a/g) a t</em>1^2/2 = 15.3 km</p>
<p>Come on man this is easy.</p>
<p>v = at
420 m/2 = 14*30</p>
<p>Then you use a constant acceleration where:</p>
<p>v naut = 420
v final= 0
a = -9.8</p>
<p>Pick an equation where the only missing variable is delta x.</p>
<p>And why is this in the SAT section? Physics isn’t on the SAT I…</p>