The 243000-lb space-shuttle orbiter touches down at about 236 mi/hr. The drag chute is deployed at 189 mi/hr, the wheel brakes a
1 answer:
Answer:
5156.37 ft
Explanation:
Given data:
weight ( W ) = 243,000 Ib
Motion of shuttle ; from 189 mi/hr to 101 mi/hr
dv/dt = -0.0002 V^2
I/v * dv/dt = -0.0002 ds
<em>Convert mi/hr to ft/s</em> ( 1 mi/hr = 1.467 ft/s)
189 mi/hr = 277.263 ft/s
101 mi/hr = 148.167 ft/s
After Integrating
In ( 148.167 / 277.263 ) = -0.0002 ( S1 - S2 )
S1 - S2 = -0.627 / -0.0002
S1 - S2 = 3135 ft/s
Now from 101 mi/hr to 35 mi/hr
dv/dt = ( - 0.0002 V^2 + 3.5 )
ds = V*dv / ( -0.0002 v^2 - 3.5 )
given : 35 mi/hr = 51.345 ft/s
101 mi/hr = 148.167 ft/s
Integrate
S3 - S2 = - In( 0.0002 v^2 + 3.5 ) / 0.0002 * 2 ]
= 1644.75 ft/s
S4 - S3 = 376.62 ft/s
attached below is the remaining part of the solution
Total distance travelled = 3135 + 1644.75 + 376.62 = 5156.37 ft
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Explanation:
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