Find Maximum Height, Time, and Duration.

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  • #1
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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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  • #2
Chi Meson
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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.
 
  • #3
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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.
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.
 
  • #4
Chi Meson
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EDIT: Nevermind, the 2.84 is the time is stops accelerating at 2 m/s^2, instead of reaching the max height.
ah haaa!
 
  • #5
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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?
 
  • #6
Chi Meson
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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.
 
  • #7
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Yes!!!! I figured it out! Thank you sooo much!
 

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