A model rocket is launched straight upward with an initial speed of 46.0 m/s. It accelerates with a constant upward acceleration of 3.00 m/s2 until its engines stop at an altitude of 130 m.
(a) What can you say about the motion of the rocket after its engines stop?
(b) What is the maximum height reached by the rocket?
(c) How long after liftoff does the rocket reach its maximum height?
(d) How long is the rocket in the air?
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a) after its engines stop the acceleration goes from 3 m/s^2 to -9.8 m/s^2 because of gravity. It starts decelerating (or you could say accelerating towards the earth).
it's velocity will be,
v^2 = vi^2 + 2*a(d-di) where vi is initial velocity and d is distance and di is initial distance
so, v^2=46^2 + 2*3(130-0)
v = 53.8 m/s at the moment the engines stop
b) i'll use v^2 = Vi^2 + 2*a(d-di) using the moment the rocket shuts it's engines off as the initial v and d point. So, v will be velocity at highest point, which means v = 0.
-(Vi^2) = 2*a(d-di)
-(Vi^2)/(2*a) + di = d
-(53.8^2)/(2*(-9.8)) + 130 = 278 m is the maximum height reached.
c) d = di + 1/2*a*t^2, d = 0 cause i'll just get the time it takes the rocket to fall from highest point which is equivalent.
(-di/(1/2(a))^(1/2) = t
(-278/(0.5*(-9.8)))^(1/2) = 7.5 seconds
d) the rocket is in the air twice as long, so 15 seconds. This is ignoring the effect of drag from the air molecules.
on a side note, I could be wrong about this but it seems strange that the rocket has an initial velocity. I believe usually in cases of rockets the initial velocity is taken as 0. To say it has an initial velocity would be like driving in a car at 46 m/s then firing the rocket out the window in the direction your driving. Obviously that's a fairly unlikely scenario.
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the rocket will continue up until the vertical speed is reduced by gravity
vertical distance till burn out:
H= v(0) t +.5 a t^2
130 = 46t + .5(3) t^2 solve for t(bo) for time at burn out
find velocity at burn out
V = V(0) + a t
v(bo) =46 + 3 t(bo)
b) What is the maximum height reached by the rocket?
max height occurs when velocity =0
V =V(bo) -.5(9.8)t
0= V(bo) -.5(9.8)t(m) solve for t(m)
max height = 130 + V(bo) t -.5(9.8) t^2 solve using t(m)
(c) How long after liftoff does the rocket reach its maximum height?
maximum height is reached when t = t(bo) +t(m)
How long is the rocket in the air?
time for rocket to fall
0 =( maxheight) -.5(9.8)t^2 solve for time for rocket to fall
total time = time to burnout + time to max height +time to fall