A rocket moves upward, starting from rest with an acceleration of 31.8 m/s2 for 3.58 s. It runs out of fuel at the end of the 3.58 s but does not stop. How high does it rise above the ground?
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first we find the height and velocity of the rocket at the end of the acceleratio phase
dist = 1/2 a t^2 = 1/2 x 31.8m/s/sx(3.58x)^2 = 203.8m
vel = a t = 31.8m/s/s x 3.58s = 113.8m/s
once the engine cuts off, the acceleration on the rocket is -9.8m/s/s, and we can find the height the rocket rises after cut off from
vf^2=v0^2 + 2ad
vf=final velocity = 0 at max ht
v0=113.8m/s at the beginning of the coasting phase
a=-9.8m/s/s
d=distance traveled in coasting phase
0=113.8^2 +2(-9.8)d => d= 660.7m
total height above the surface = 864.4m
dist = 1/2 a t^2 = 1/2 x 31.8m/s/sx(3.58x)^2 = 203.8m
vel = a t = 31.8m/s/s x 3.58s = 113.8m/s
once the engine cuts off, the acceleration on the rocket is -9.8m/s/s, and we can find the height the rocket rises after cut off from
vf^2=v0^2 + 2ad
vf=final velocity = 0 at max ht
v0=113.8m/s at the beginning of the coasting phase
a=-9.8m/s/s
d=distance traveled in coasting phase
0=113.8^2 +2(-9.8)d => d= 660.7m
total height above the surface = 864.4m
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864 meters