A rocket rises vertically, from rest, with an acceleration of 3.2 m/s^2 until it

Question

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s^2 until it runs out of fuel at an altitude of 1500 m. After this point, its acceleration is that of gravity, downward.
What is the velocity of the rocket when it runs out of fuel?
Express your answer using two significant figures.
v_{1500 <units>m</units>} =


{ m m/s}
submitshow answer

Part B
How long does it take to reach this point?
Express your answer using two significant figures.
t_{1500 <units>m</units>} =


{ m s}
submitshow answer

Part C
What maximum altitude does the rocket reach?
Express your answer using two significant figures.
y_{max} =


{ m m}
submitshow answer

Part D
How much time (total) does it take to reach maximum altitude?
Express your answer using two significant figures.
t =


{ m s}
submitshow answer

Part E
With what velocity does the rocket strike the Earth?
Express your answer using two significant figures.
v_{ m strike} =


{ m m/s}
submitshow answer

Part F
How long (total) is it in the air?
Express your answer using two significant figures.
t_{ m total} =


{ m s}
submitshow answer

Explanation / Answer

For this multi-part problem, you will need to know the three basic acceleration equations and use the appropriate one depending on the data given: d = v't + 0.5at^2 v" = v' + at v"^2 - v'^2 = 2ad where d = distance (meters) t = time (seconds) a = acceleration (meters/sec^2) v' = original velocity (meters/sec) v" = final velocity (meters/sec) a) All you have is the acceleration (3.2), original velocity (0), and distance (1400), so use the 3rd equation v"^2 - v'^2 = 2ad --> v"^2 - 0 = 2(3.2)(1400) --> v" = sqrt8960 = 94.66m/s b) You now have the values to plug into the 2nd equation to solve for t v" = v' + at --> t = (v" - v')/a = (94.66 - 0)/3.2 = 29.58s c) Maximum altitude will be 1400m + the distance d the rocket continues to "coast" upwards until gravitational accelearation (-9.8) brings it to a complete stop. Use the 3rd equation to figure this additional distance, but now v' = 94.66 and v" = 0 v"^2 - v'^2 = 2ad --> d = (v"^2 - v'^2)/2a = (0 - 94.66^2)/2(-9.8) = -8960/19.6 = 457.14m Max altitude = 1400 + 457.14 = 1857.14m d) Total time for maximum altitude will be the time from part b (29.58) + the time t to decelerate through the 457.14 m from part c. Using the 2nd equation, and acceleration of -9.8: v" = v' + at --> t = (v" - v')/a = (0 - 94.66)/-9.8 = 9.66s So the total climb time = 29.58 + 9.66 = 39.24s e) Presuming no drag or terminal velocity, you now have the data to plug into the 3rd equation again. v' = 0, a = -9.8, and d = -1857.14 (negative because you're falling downwards) v"^2 - v'^2 = 2ad --> v"^2 = 2ad + v'^2 = 2(-9.8)(-1857.14) + 0 --> v" = sqrt36399.94 = -190.79m/s (again, negative because velocity is downwards) f) Total air time will equal time rising (29.58) + time falling t v" = v' + at --> t = (v" - v')/a = (-190.79 - 0)/-9.8 = 19.47s So total time = 29.58 + 19.47 = 49.05s

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