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?

Question

Henry · Answer

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.