R My EEE EEE x M Ch 2.7 Derivs and+R x Stewart Calculus: Early x Assignments x M
ID: 1527220 • Letter: R
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R My EEE EEE x M Ch 2.7 Derivs and+R x Stewart Calculus: Early x Assignments x Mastering Physics: Wee x C During Launches, Rock x C session masteringphysics.com 75109036&offset; next Physics 2 Winter Quarter 2017 Help I Close Resources Week Long etison e previous l 4 of 13 l next SWB-Rocket-jettison Part A During launches, rockets often discard unneeded parts. How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration? A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.05 m/s When it is 260 m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored). Submit My Answers Give Up Part B What total distance did the canister travel between its release and its crash onto the launch pad? Submit My Answers Give Up Provide Feedback ContinueExplanation / Answer
here,
acceleration, a = 3.05 m/s^2
height , h = 260 m
from third eqn of motion,
velocity, v = sqrt(2*g*h)
velocity, v = sqrt(2*9.81*260)
velocity, v = 71.423 m/s
Time taken by cannister to reach ground
-260 = 71.423 *t - 0.5 * 9.81 * t^2
time, t = 17.577 s
Part a:
height the rocket reaches in the given time...
h = 260 + 71.423*(17.577) + 0.5*(3.05)(17.577)^2
h = 1986.552 m
part b:
time for rise of cannister when blasted away from rocket
t = v/g
t = 71.423/9.81
t = 7.3 s
additional height gained , ha = 3.05 * 7.3 = 22.265 m
total distance to fall, hf = 22.265 + 260 = 282.265 m
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