Posted by **matin** on Wednesday, May 9, 2012 at 8:14pm.

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 ??

- physics -
**Elena**, Thursday, May 10, 2012 at 10:08am
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=vₒ-g•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.

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