Question

Early test flights for the space shuttle used a "glider" (mass of 940 kg including pilot). After a horizontal launch at 450 km/h at a height of 3900 mm , the glider eventually landed at a speed of 230 km/h.

A) What would its landing speed have been in the absence of air resistance?

B) What was the average force of air resistance exerted on it if it came in at a constant glide angle of 14 degrees to the Earth's surface?

Answer 1

a)

Using conservation of energy in the absence of air resistance,

increase in kinetic energy of the glider = decrease in potential energy of the glider

mvmu - = mgh

2 (450 * 2 . 18 = 9.8 * 3900

v = 303.422 m/s = 303.422 * = 1092.32 km/h

b)

By work energy theorem,

Work done by gravity + Work done by air resistance = Change in kinetic energy

mv2 mu? mgh + Wair = 3-2

, 940 (450+ 12 940 + (230 ) 940 +9.8 + 3900 + Wor=-

Wair = -3.05 * 107 J

This implies displacement is opposite in direction to the air resistance force.

If the air resistance is acting at a glide angle $\Theta$ ,

3.05 * 107 = Fair * ** sin(0)

Fair = 3.05 * 10'* sin(14°) ? = 1891.95 N 3900