Find Maximum Height, Time, and Duration.

In summary, using the GUESS method, we can find that the rocket reaches a max height of 308.16 m and it takes 8.52 s to reach this height. The total time the rocket is in the air is 16.45 s. The equation X = V_o t + 1/2at^2 can be used to find t1, the time it takes for the rocket to stop accelerating at 2 m/s^2. Then, using the equation V = V_o + at, we can find t2, the time it takes for the rocket to reach its maximum height. The final speed at the top is 0 m/s. The total time for the rocket to lose all its
  • #1
petern
89
0
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
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
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.
 
  • #4
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!
 
  • #5
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
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
Yes! I figured it out! Thank you sooo much!
 

1. What is the formula for finding the maximum height of an object?

The formula for finding the maximum height of an object is h = (v2sin2θ)/(2g), where h is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

2. How do you calculate the time it takes for an object to reach its maximum height?

The formula for calculating the time it takes an object to reach its maximum height is t = v sin θ/g, where t is the time, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

3. What factors can affect the maximum height of an object?

The maximum height of an object can be affected by the initial velocity, the angle of projection, and the acceleration due to gravity. Other factors such as air resistance and wind can also have an impact.

4. Is there a way to find the duration of an object's flight?

Yes, the duration of an object's flight can be calculated using the formula t = 2v sin θ/g, where t is the duration, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

5. Can the maximum height, time, and duration of an object be calculated for any type of motion?

Yes, the formulas for finding the maximum height, time, and duration of an object are applicable to any type of motion as long as the initial velocity, angle of projection, and acceleration due to gravity are known.

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