physics
posted by matin .
A rocket is launched vertically from Earth's surface with a velocity of 3.4 km/s. How high does it go:
From Earth's centre?
From Earth's surface?
how can this be done without any time or force mentioned ??

vₒ= 3.4 km/s= 3400 m/s.
The problem may be solved by several methods.
The first method (kinematics)
The height from the Earth's surface
h =vₒ•t g•t²/2,
v=vₒg•t.
At the top point v=0, => vₒ=g•t, t = vₒ/g,
Substitute it in the equation for h and obtain
h=vₒ²/2•g =(3400 )²/2•9.8 ≈ 5.9•10^5 m. =590 km.
The second method (the law of conservation of energy)
KE = ΔPE.
m•v²/2 = m•g•h,
h = vₒ²/2•g =….
The Earth’s radius is R ≈ 6378 km.
The distance from Earth's centre is
H = 6378 + 590 = 6968 km.