Posted by Lindsay on Sunday, September 30, 2007 at 12:34pm.
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
For part a, my book says the answer is 310 m. I'm still really confused on how they got this answer.
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