a) First, calculate the velocity Vmax attained during acceleration at rate a. (They already tell you the altitude there, H = 150 m). Then compute how huch higher it "coasts" before reaching maximum altitude.
Vmax = sqrt(2aH)= 24.5 m/s
To have the velocity decrease to zero, the additional time of flight t' is given by
g t' = 24.5 m/s, so
t' = 24.5/9.8 = 2.5 s
The average speed while coasting to zero velocity is Vmax/2 = 12.25 m/s.
Additionl altitude gained = (2.5)(12.25) = 30.63 m
Maximum altitude = Hmax = 150 + 30.63 = 180.63 m
b) it attains maximu height 2.5 seconds after accelerqation stopped. We already calculated that. Add that to the tiome spent accelerating for the total time after launch. The time t spent accelerating is given by
(1/2) a t^2 = 150 m
t = sqrt (2*150/a)= 12.25 s
Total time to reach maximum altitude - t + t' = ?
c) Add to the last answer the time that it takes to fall back down. Call this time t"
(1/2) g t"^2 = Hmax
Solve for t" and add it to the time in the previous answer
A model rocket is launched straight upward with an initial speed of 50 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.
a) What is the max. height reached by the rocket?
b) When does the rocket reach max. height?
c) How long is the rocket in the air?
I really need help with which equation to use for this problem. I'm unsure of even where to begin...!
2 answers
For part a, my book says the answer is 310 m. I'm still really confused on how they got this answer.
1 answer