Find Maximum Height, Time, and Duration.

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A model rocket is launched with an initial speed of 50 m/s and a constant upward acceleration of 2 m/s² until it reaches an altitude of 150 m. The maximum height achieved by the rocket is 308.16 m, reached at 8.52 seconds. The total time the rocket remains in the air is 16.45 seconds. The discussion highlights the importance of correctly applying kinematic equations to solve for time and height in projectile motion problems. Participants clarified the steps needed to find the time intervals for both the powered ascent and the subsequent free flight.
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Problem: A model rocket is launched straight upward with an initial speed of 50 m/s. It accelerates with a constant upward acceleration of 2 m/s^2 until the engines stop at an altitude of 150 m. (Use GUESS method, NO working equation.)
a) What is the max height reached by the rocket? Answer: delta x = 308.16 m
b) When does the rocket reach max height? Answer: t-up = 8.52 s
c) How long is the rocket in the air? Answer: t = 16.45 s


I can't figure out this problem at all. I know that I have to use the equation X = X_o t + 1/2at^2. The problem is, I don't know what t is and it's what's being asked in b and c. The equations V = V_o + at doesn't work either because I'm missing t. The equation V^2 = V_o^2 + 2a (delta)X doesn't work either because I'm missing the X which is asked for in A. Please help.
 
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This problem has two parts.

The first part is the firing of the rocket where the acceleration is +2 m/s^2. There is an initial velocity and a vertical displacement given. The formula is slightly wrong, it should be X = V_o t + 1/2at^2. Use it to find t1.

Part two, the rocket is still going up (what is the speed when the engine quits) but now the acceleration is that of gravity. How much time to reach the top (what is final speed at the very top?) Find t2, the total time-up.
 
Chi Meson said:
This problem has two parts.

The first part is the firing of the rocket where the acceleration is +2 m/s^2. There is an initial velocity and a vertical displacement given. The formula is slightly wrong, it should be X = V_o t + 1/2at^2. Use it to find t1.

Part two, the rocket is still going up (what is the speed when the engine quits) but now the acceleration is that of gravity. How much time to reach the top (what is final speed at the very top?) Find t2, the total time-up.

OK, I try 150 = 50t + (1/2)(2)(t^2) and get t = 2.84 s. That is wrong because according to b), it's 8.52 s. What am I doing wrong?

EDIT: Nevermind, the 2.84 is the time is stops accelerating at 2 m/s^2, instead of reaching the max height.
 
petern said:
EDIT: Nevermind, the 2.84 is the time is stops accelerating at 2 m/s^2, instead of reaching the max height.

ah haaa!
 
OK, so I get t1 as 2.84 and t2 as 4.98 which equals 7.82 but ti's still not 8.52 as given in b). What am I doing wrong?
 
It's not 4.98 s . What is its speed when the rocket quits? 50m/s + at ?

How long does it take to lose all this speed at a rate of 9.8 m/s per second? I'm getting 8.52 total.
 
Yes! I figured it out! Thank you sooo much!
 

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