Initial speed of projectile from the surface of earth, v=αve Where, α is constant and ve is escape velocity of projectile.. Radius of earth, RE=6400‌km =6.4×106m Maximum height attained by the projectile, h=800‌km=8×105m We know that, escape velocity on the surface of earth, ve‌=I1.2‌km‌/‌s ‌=11.2×103m/s ‌v=αve ‌v=α×11.2×103.....(i) According to given situation, applying the law of conservation of energy ‌‌