Early test flights for the space shuttle used a glider

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The discussion focuses on calculating the landing speed of a glider used in early space shuttle test flights, specifically in the absence of air resistance. The initial calculations attempted to determine the landing speed based on energy conservation principles, but the user encountered errors in their approach. They also sought assistance in calculating the average force of air resistance, considering a glide angle of 15 degrees. The conversation highlights confusion around applying the equations of motion and the impact of air resistance on the glider's descent. Clarification on these physics concepts is needed to resolve the calculations accurately.
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Homework Statement


Early test flights for the space shuttle used a "glider" (mass of 940 kg including pilot). After a horizontal launch at 470 km/h at a height of 4000 m, the glider eventually landed at a speed of 210 km/h . What would its landing speed have been in the absence of air resistance(in km/h)? What was the average force of air resistance exerted on it if it came in at a constant glide angle of 15 degrees to the Earth's surface?

Homework Equations


Ki + Ui = Kf + Uf
.5mv^{2}+mgh=.5mv^{2}

The Attempt at a Solution


I'm not so sure about the second question but here's what I did for the first one.
I changed the units so that the answer would be in km/h.
g=35.28 km/h
h=4 km
v^{2}=[2(.5mv^{2}+mgh)]/m
v^{2}=[2(.5(940)(470)^{2}+(940)(35.28)(4)]=221182.24
v=470km/h
Unfortunately, this answer is incorrect. Any help would be greatly appreciated.
 
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g is acceleration? The units could be kilometers per hour squared, rather than kilometers per hour.
 


In the absence of air resistance, the horizontal velocity will remain as it is.
Find its velocity when if falls through 4 km. The resultant of this velocity and the horizontal velocity will give the landing velocity.
 


Thanks. I got the first part. As for the second question, I'm not quite sure how to go about it. I think I'm suppose to use F=ma. So I caculated the acceleration.
a=(v^{2}-v^{2}_{0})/(2x)
a=(210^{2}-470^{2})/((2)(4))=-22100

F=ma=(940)(-22100)=-2077400

But that only gets me the force of the gilder and doesn't include friction. I was also thinking that I should get the sum of forces.So
F_{gilder}-F_{friction}=ma
I don't think that's quite right either. There suppose to be a 15 degree angle but I don't know where it goes. Can somebody help me?
 
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