Determine the launch elevation and maximum range: the test vehicle should obtain a burnout velocity of 7200 m/s at 180 km altitude (rburnout = 6558 km). what would the flight path angle (fburnout) be?

Flight path angle= 15.12°, maximum range= 5.29× 10*6 km

Explanation:

u= 7200m/s, H= 180km= 180000m

Recall that

Maximum height, H= (u*2sin*2∆)/2g

180000= (7200×7200sin*2∆)/2×9.8

(18000×2×98)/7200×7200= sin*2∆

Sin∆= 0.2609

∆= 15.12°

Maximum range, R= u*2/g

(7200×7200)/9.8

= 5289795.92km

= 5.29× 10*6 km

c. newton's third law of motion