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Chapter 4: Problem 134

A model rocket is fired vertically from rest. It has a net acceleration of \(17.5 \mathrm{m} / \mathrm{s}^{2} .\) After \(1.5 \mathrm{s}\), its fuel is exhausted and its only acceleration is that due to gravity. (a) Ignoring air resistance, how high does the rocket travel? (b) How long after liftoff does the rocket return to the ground?

Short Answer

Expert verified
The rocket's maximum height is 54.5191 meters, and the total time it takes for the rocket to return to the ground is 4.8379 seconds.

Step by step solution

01

Part 1: Upward motion with net acceleration

During the first \(1.5 \mathrm{s}\), the rocket has a net acceleration of \(a = 17.5 \mathrm{m}/\mathrm{s}^2\). We can find the final velocity (\(v_1\)) and the height (\(h_1\)) at the end of this interval using the equations: \(v_1 = u+at\) \(h_1 = ut+\frac{1}{2}at^2\) where \(u = 0\) (initial velocity) and \(t = 1.5 \mathrm{s}\) (time). Calculating \(v_1\) and \(h_1\): \(v_1 = 0+(17.5)(1.5) = 26.25 \mathrm{m}/\mathrm{s}\) \(h_1 = (0)(1.5)+\frac{1}{2}(17.5)(1.5)^2 = 19.6875 \mathrm{m}\)
02

Part 2: Upward motion with acceleration due to gravity

Now the rocket continues moving upward with an initial velocity of \(v_1 = 26.25 \mathrm{m}/\mathrm{s}\), but with acceleration \(a = -9.8 \mathrm{m}/\mathrm{s}^2\) due to gravity. We will use the equation: \(v_{2}^2 = v_{1}^2 + 2ah_2\) to find the height (\(h_2\)) at which the rocket's velocity becomes \(0\): \(0^2 = (26.25)^2 + 2(-9.8)h_2\) Solving for \(h_2\): \(h_2 = \frac{(26.25)^2}{2(9.8)} = 34.8316\mathrm{m}\)
03

Part 3: Downward motion with acceleration due to gravity

The rocket now falls back to the ground from a height \(h_T = h_1 + h_2 = 19.6875 + 34.8316 = 54.5191 \mathrm{m}\) with acceleration \(a = -9.8 \mathrm{m}/\mathrm{s}^2\). To find the time (\(t_3\)) it takes to fall back to the ground, we will use the equation: \(h_T = ut_3+\frac{1}{2}a{t_3}^2\) As the initial velocity at the top is \(0\), so \(u = 0\). Now, we can solve for \(t_3\): \(54.5191 = 0 + \frac{1}{2}(-9.8){t_3}^2\) Solving for \(t_3\): \(t_3 = \sqrt{\frac{2(54.5191)}{9.8}} = 3.3379 \mathrm{s}\)
04

Final answers

(a) The rocket's maximum height is \(h_T = 54.5191 \mathrm{m}\). (b) The total time it takes for the rocket to return to the ground is \(t_{total} = t + t_3 = 1.5 + 3.3379 = 4.8379 \mathrm{s}\).

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