Question 2.10: A ROCKET GOES BALLISTIC GOAL Solve a problem involving a pow...
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 \mathrm{~m} / \mathrm{s}^{2}. It runs out of fuel at the end of 4.00 \mathrm{~s} 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 4.00 \mathrm{~s}. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes on the ground.
STRATEGY Take y=0 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 29.4 \mathrm{~m} / \mathrm{s}^{2}, and we can use the kinematic equations with constant acceleration a 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 9.80 \mathrm{~m} / \mathrm{s}^{2}, with initial velocity and position given by the results of phase 1. Apply the kinematic equations for free fall.
(a) Phase 1: Find the rocket’s velocity and position after 4.00 \mathrm{~s}.
Write the velocity and position kinematic equations:
(1) v=v_{0}+a t
(2) \Delta y=y-y_{0}=v_{0} t+\frac{1}{2} a t^{2}
Adapt these equations to phase 1 , substituting a=29.4 \mathrm{~m} / \mathrm{s}^{2}, v_{0}=0, and y_{0}=0 :
(3) v=\left(29.4 \mathrm{~m} / \mathrm{s}^{2}\right) t
(4) y={ }_{2}^{1}\left(29.4 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}=\left(14.7 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}
Substitute t=4.00 \mathrm{~s} into Equations (3) and (4) to find the rocket’s velocity v and position y at the time of burnout.
These will be called v_{b} and y_{b}, respectively.
v_{b}=118 \mathrm{~m} / \mathrm{s} and y_{b}=235 \mathrm{~m}
(b) Phase 2: Find the maximum height the rocket attains.
Adapt Equations (1) and (2) to phase 2, substituting a=9.8 \mathrm{~m} / \mathrm{s}^{2}, v_{0}=v_{b}=118 \mathrm{~m} / \mathrm{s}, and y_{0}=y_{b}=235 \mathrm{~m} :
(5) v=\left(-9.8 \mathrm{~m} / \mathrm{s}^{2}\right) t+118 \mathrm{~m} / \mathrm{s}
(6) y=235 \mathrm{~m}+(118 \mathrm{~m} / \mathrm{s}) t-\left(490 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}
Substitute v=0 (the rocket’s velocity at maximum height) in Equation (5) to get the time it takes the rocket to reach its maximum height:
0=\left(-9.8 \mathrm{~m} / \mathrm{s}^{2}\right) 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}
Substitute t=12.0 \mathrm{~s} into Equation (6) to find the rocket’s maximum height:
\begin{aligned}y_{\max } &=235 \mathrm{~m}+(118 \mathrm{~m} / \mathrm{s})(12.0 \mathrm{~s})-\left(4.90 \mathrm{~m} / \mathrm{s}^{2}\right)(12.0 \mathrm{~s})^{2} \\&=945 \mathrm{~m}\end{aligned}
(c) Phase 2: Find the velocity of the rocket just prior to impact.
Find the time to impact by setting y=0 in Equation (6) and using the quadratic formula:
\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}
Substitute this value of t into Equation (5):
v =\left(-9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(25.9 \mathrm{~s})+118 \mathrm{~m} / \mathrm{s}=-136 \mathrm{~m} / \mathrm{s}
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.