AP EAPCET 04-Jul-2022 Shift 1 Paper

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
‌‌

1

2

mv2=mgh
‌⇒‌‌v2=2gh⇒(αve)2=2gh