Thrust phase:
v = a t
v = 17 t
v at 20 s = 17 * 20 = 340 m/s (Vi for next phase)
h = (1/2) a t^2 = (1/2)(17)(400) = 3,400 m
coast up phase:
v = Vi - a t
v = 0 at top
0 = 340 - 9.81 t
t = 34.7 seconds coasting up
now 54.7 seconds aloft
how high?
h = 3,400 + Vi t - 4.9 t^2
h = 3,400 +340(34.7) -4.9(34.7)^2
= 3,400 + 11,784 - 5,886
= 9,298 m made like 9 km, not 12
now how long to fall
0 = 9.298 - 4.9 t^2
t = 43.6 s
so 54.7 + 43.6 = 98.3 seconds total aloft
You have designed a rocket to be used to sample the local atmosphere for pollution. It is fired vertically with a constant upward acceleration of 17 m/s2. After 20 s, the engine shuts off and the rocket continues rising (in freefall) for a while. (Neglect any effects due to air resistance.) The rocket eventually stops rising and then falls back to the ground. You want to get a sample of air that is 12 km above the ground.
(a) What is the highest point your rocket reaches?
(b) Determine the total time the rocket is in the air.
(c) Find the speed of the rocket just before it hits the ground.
2 answers
(c) v = sqrt (2 g h)