stage one:
h = 45 t + (1/2)(1.5) t^2 = 160
solve for t
.75 t^2 + 45 t - 160 = 0
t = 3.367 seconds
v = 45 + 1.5 t = 45 + 1.5(3.367) = 50.05 m/s
stage 2
v = Vi - g t
at top v = 0
9.81 t = 50.05
t = 5.1 seconds drifting up
h = 160 + 50.05(5.1)-4.9(5.1)^2
= 160 + 255 - 127
= 287 meters at the top
time so far = 3.4+5.1 = 8.5 seconds
stage 3, ignominious fall to ground
h = 287
so
287 = 4.9 t^2
t = 7.65 second fall
so
time in air = 8.5 + 7.7 = 16.2 seconds
A model rocket is launched straight upward with an initial speed of 45.0 m/s. It accelerates with a constant upward acceleration of 1.50 m/s2 until its engines stop at an altitude of 160 m.
A. what is the max height reached by the rocket?
B. How long after liftoff does the rocket reach its maximum height?
c. How long is the rocket in the air?
1 answer
1 answer