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professordad
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Homework Statement
A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s^2 until its engines stop at an altitude of 150 m. How long after lift off does the rocket reach the maximum height?
from College Physics by Serway and Faughn
Homework Equations
(1) v = v_0 + at where v denotes velocity, v_0 denotes initial velocity, a denotes acceleration, and t denotes time in seconds.
(2) x = v_0t + \frac{1}{2}at^2 where x denotes the displacement
(3) v^2 = v_0^2 + 2ax
The Attempt at a Solution
Ok so for this problem I found 2 ways to do it. One of them gives the correct answer (in the back of the book) and the other is off by a little.
The method that gives the correct answer:
So let the velocity of the rocket at the point where the acceleration stops be v_1 and the velocity of the rocket at the highest point be v_2. Let the time of the rocket's acceleration be t_1 and the time of no rocket's acceleration be t_2. Note that at the highest point, the velocity is 0, so v_2 = 0.
Now we find v_1. From equation (3), v_1^2 - 50.0^2 = 2 \cdot 2.00 \cdot 150 so v_1 = 55.68 m/s.
From equation (1), we have v_1 = 50.0 + at_1, so 55.68 = 50.0 + 2.00 \cdot t_1 and t_1 = 2.84 sec.
From equation (1) again, we have v_2 = v_1 + at_2, so -55.68 = -9.8t and t_2 = 5.68 sec.
The answer is \boxed{8.52} sec.
Ok so here's the other method that I tried but it didn't give the right answer:
Let d_2 denote the distance that the rocket travels after it stops accelerating itself. Then, by (3), 0 = 55.68^2 - 2 \cdot 9.8 \cdot x, so x = 158.
So by (2), x = 158 = 55.68 \cdot t_2 - 4.9t_2^2.
Solving the quadratic gives t_2 = 5.49 sec or t = 5.87 sec.
But this doesn't give the right answer :( , help?
