A ROCKET GOES BALLISTIC GOAL Solve a problem involving a powered ascent followed by free-fall motion. PROBLEM A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s². It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height

Question

A ROCKET GOES BALLISTIC

GOAL Solve a problem involving a powered ascent followed by free-fall motion.

PROBLEM A rocket moves straight upward, starting from rest with an acceleration of [latex]+29.4 \mathrm{~m} / \mathrm{s}^{2}[/latex]. It runs out of fuel at the end of [latex]4.00 \mathrm{~s}[/latex] and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket's veloc ity and position at the end of [latex]4.00 \mathrm{~s}[/latex]. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes on the ground.

STRATEGY Take [latex]y=0[/latex] at the launch point and y positive upward, as in Figure 2.21. The problem con sists of two phases. In phase 1 the rocket has a net upward acceleration of [latex]29.4 \mathrm{~m} / \mathrm{s}^{2}[/latex], and we can use the kinematic equations with constant acceleration [latex]a[/latex] to find the height and velocity of the rocket at the end of phase 1 , when the fuel is burned up. In phase 2 the rocket is in free fall and has an acceleration of [latex]9.80 \mathrm{~m} / \mathrm{s}^{2}[/latex], with initial velocity and position given by the results of phase 1. Apply the kinematic equations for free fall.

Accepted Answer

(a) Phase 1: Find the rocket's velocity and position after [latex]4.00 \mathrm{~s}[/latex].

Write the velocity and position kinematic equations:

(1) [latex]v=v_{0}+a t[/latex]

(2) [latex]\Delta y=y-y_{0}=v_{0} t+\frac{1}{2} a t^{2}[/latex]

Adapt these equations to phase 1 , substituting [latex]a=29.4 \mathrm{~m} / \mathrm{s}^{2}, v_{0}=0[/latex], and [latex]y_{0}=0[/latex] :

(3) [latex]v=\left(29.4 \mathrm{~m} / \mathrm{s}^{2} ight) t[/latex]

(4) [latex]y={ }_{2}^{1}\left(29.4 \mathrm{~m} / \mathrm{s}^{2} ight) t^{2}=\left(14.7 \mathrm{~m} / \mathrm{s}^{2} ight) t^{2}[/latex]

Substitute [latex]t=4.00 \mathrm{~s}[/latex] into Equations (3) and (4) to find the rocket's velocity [latex]v[/latex] and position [latex]y[/latex] at the time of burnout.
These will be called [latex]v_{b}[/latex] and [latex]y_{b}[/latex], respectively.

[latex]v_{b}=118 \mathrm{~m} / \mathrm{s}[/latex] and [latex]y_{b}=235 \mathrm{~m}[/latex]

(b) Phase 2: Find the maximum height the rocket attains.

Adapt Equations (1) and (2) to phase 2, substituting [latex]a=9.8 \mathrm{~m} / \mathrm{s}^{2}, v_{0}=v_{b}=118 \mathrm{~m} / \mathrm{s}[/latex], and [latex]y_{0}=y_{b}=235 \mathrm{~m}[/latex] :

(5) [latex]v=\left(-9.8 \mathrm{~m} / \mathrm{s}^{2} ight) t+118 \mathrm{~m} / \mathrm{s}[/latex]

(6) [latex]y=235 \mathrm{~m}+(118 \mathrm{~m} / \mathrm{s}) t-\left(490 \mathrm{~m} / \mathrm{s}^{2} ight) t^{2}[/latex]

Substitute [latex]v=0[/latex] (the rocket's velocity at maximum height) in Equation (5) to get the time it takes the rocket to reach its maximum height:

[latex]0=\left(-9.8 \mathrm{~m} / \mathrm{s}^{2} ight) t+118 \mathrm{~m} / \mathrm{s} \quad → \quad t=\frac{118 \mathrm{~m} / \mathrm{s}}{980 \mathrm{~m} / \mathrm{s}^{2}}=12.0 \mathrm{~s}[/latex]

Substitute [latex]t=12.0 \mathrm{~s}[/latex] into Equation (6) to find the rocket's maximum height:

[latex]\begin{aligned}y_{\max } &=235 \mathrm{~m}+(118 \mathrm{~m} / \mathrm{s})(12.0 \mathrm{~s})-\left(4.90 \mathrm{~m} / \mathrm{s}^{2} ight)(12.0 \mathrm{~s})^{2} \&=945 \mathrm{~m}\end{aligned}[/latex]

(c) Phase 2: Find the velocity of the rocket just prior to impact.

Find the time to impact by setting [latex]y=0[/latex] in Equation (6) and using the quadratic formula:

[latex]\begin{aligned}0 &=235 \mathrm{~m}+(118 \mathrm{~m} / \mathrm{s}) t-(4.90 \mathrm{~m} / \mathrm{s}) t^{2} \t &=25.9 \mathrm{~s} \end{aligned}[/latex]

Substitute this value of [latex]t[/latex] into Equation (5):

[latex]v =\left(-9.80 \mathrm{~m} / \mathrm{s}^{2} ight)(25.9 \mathrm{~s})+118 \mathrm{~m} / \mathrm{s}=-136 \mathrm{~m} / \mathrm{s}[/latex]

REMARKS You may think that it is more natural to break this problem into three phases, with the second phase ending at the maximum height and the third phase a free fall from maximum height to the ground. Although this approach gives the correct answer, it's an unnecessary complication. Two phases are sufficient, one for each different acceleration.

Question 2.10: A ROCKET GOES BALLISTIC GOAL Solve a problem involving a pow...

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