8. 0.211 points l Previous Answers Tipler62.PO79 D My Notes Ask Your Teacher You

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8. 0.211 points l Previous Answers Tipler62.PO79 D My Notes Ask Your Teacher 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 19 m/s2. After 25 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 20 km above the ground. (a) What is the highest point your rocket reaches? X km Did you reach your height goal? Yes O No If not, what would you change so that the rocket reaches 20 km? This answer has not been graded yet. (b) Determine the total time the rocket is in the air. 50.0 (c) Find the speed of the rocket just before it hits the ground m/s

Explanation / Answer

1. from the given data:
calculate the distance to where the engine JUST shuts off.
s = ut + ½at²
s = 0*25 + 0.5*19*25^2
s = 5937.5 m

Calculate the peak. (from the point where the engine shuts off)
v = u + at
v = 0 + 19*25
v = 475 m/s

v² = u² + 2as
0² = 475² + 2*(-9.81)s
2*9.81s = 475²
s = 11499.74 m

Highest point = 11499.74+5937.5
=17437.24m
=17.43724 km

THUS it is not possible to collect a sample of air at 20km above ground.

time taken to rise 11499.74m
v = u + at
0 = 475+ -9.81t
t =48.42 s

TOTAL time = 48.42+ 25 = 73.42 s

final velocity when rocket hits the ground=
v² = u² + 2as
v² = 0² + 2(9.81)*(17437.24)
v=585 m/s

2.From the given data:
We have...
a)the velocity as a function of time during the interval.
v(t) = 0.1*t^2/2 + c
8.8 = (0.1*(9.8)^2)/2 + c
c = 3.998
v(t) = 0.1*t^2/2 + 3.998
v(t) = 0.05*t^2 + 3.998

b) the position as a function of time during the interval.

v(t) = 0.05*t^2 + 3.998
x(t) =0.05*(t^3)/3+3.998*t+C
5.5=(0.05*(9.8^3)/3)+3.998*9.8+C
C=-49.36
x(t) =0.05*(t^3)/3+3.998*t-49.36

c)
avg. velocity v = change in position / change in time
v = (((0.05*(9.8^3)/3)+3.998*9.8-49.360)-((0.05*(0^3)/3)+3.998*0-49.36))/(9.8-0)
v= 5.598 m/s
the average of the instantaneous velocities at the start and ending times are not equal

3. the gravitational potential energy equation, by integration
GPE = -m*integral(g(x) dr)

g(x) = g0*R^2/x^2, notice this ONLY applies outside the surface

GPE(x) = m*g0*R^2*(1/x)

Evaluate GPE at both locations
GPE(3*R) = m*g0*R^2/(3*R) = 5*m*g0*R
GPE(R) = m*g0*R

Find the change in GPE, which will equal the increase in KE. Since it started at rest, this will equal the KE just before impact.

dGPE = KE_impact = GPE(3*R) - GPE(R)

KE_impact = 2*m*g0*R

Neglect relativistic effects and find the corresponding velocity:
KE_impact =1/2*m*v^2

v=sqrt(2*KE_impact/m)

v = sqrt(4*g0*R)
v=sqrt(4*1.63*3200)=144.44 m/s = 0.14444km/s

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