A model rocket is launched straight upward with an initial speed of 46.4 m/s. It accelerates with a constant upward acceleration of 2.18 m/s2 until its engines stop at an altitude of 138 m. What is the maximum height reached by the rocket? How long after lift off does the rocket reach its maximum height?How long is the rocket in the air?
a. V^2=Vo^2 + 2a*h=(46.4)^2 - 4.36*138 =
1551.28.
V = 39.4 m/s. at shut-off.
h max = 138m + -(Vo^2)/2g = 138 + (39.4^2)/19.6 = 217 m. Above gnd.
b. V = Vo + a*T1 = 39.4 m/s.
46.4 - 2.18T1 = 39.4.
-2.18T1 = - 7.
T1 = 3.21 s.
h = Vo*T2 + 0.5g*T2^2 = 217-138.
39.4T2 - 4.9T2^2 = 79.
-4.9T2^2 + 39.4T2 - 79 = 0.
Use Quadratic Formula.
T2 = 4.22 s.
T1+T2 = 3.21 + 4.22 = 7.43 s. To reach
max ht.
c. h = 0.5g*T3^2 = 217.
4.9T3^2 = 217.
T3^2 = 44.3.
T3 = 6.66 s. = Fall time.
T1+T2+T3 = 3.21 + 4.22 + 6.66 = 14.1 s.
= Time in air.
To find the maximum height reached by the rocket, we need to determine the time it takes for the rocket to reach that height. We can use the following equation of motion:
vf^2 = vi^2 + 2ad
where:
vf = final velocity of the rocket (which is zero at maximum height)
vi = initial velocity of the rocket (46.4 m/s)
a = acceleration of the rocket (2.18 m/s^2)
d = the displacement or change in height (138 m)
By substituting the known values into the equation, we can solve for the time t it takes for the rocket to reach its maximum height:
0 = (46.4 m/s)^2 + 2 * 2.18 m/s^2 * d
Rearranging the equation:
2 * 2.18 m/s^2 * d = -(46.4 m/s)^2
Simplifying:
d = -[(46.4 m/s)^2] / [2 * 2.18 m/s^2]
Plugging in the values and calculating:
d ≈ -1011.93 m
Here, the negative sign indicates that the displacement is in the opposite direction to the initial motion (upward). However, since we are interested in the magnitude of the displacement (the height), we can ignore the negative sign.
Therefore, the maximum height reached by the rocket is approximately 1011.93 meters.
To find the time at which the rocket reaches its maximum height, we can now use the following equation of motion:
vf = vi + at
where:
vf = final velocity (zero at maximum height)
vi = initial velocity (46.4 m/s)
a = acceleration (2.18 m/s^2)
t = time
By rearranging the equation:
t = (vf - vi)/a
Substituting the values:
t = (0 - 46.4 m/s) / 2.18 m/s^2
Calculating:
t ≈ -21.28 s
Again, the negative sign is due to the opposite direction of motion. However, we can ignore it since we are interested in the magnitude of time.
Therefore, the rocket reaches its maximum height after approximately 21.28 seconds.
To find the total time the rocket is in the air, we need to consider the time it takes the rocket to reach its maximum height and double it.
Total time = 2 * time to reach maximum height
Total time = 2 * 21.28 s
Calculating:
Total time ≈ 42.56 s
Therefore, the rocket is in the air for approximately 42.56 seconds.