Posted by matin on Wednesday, May 9, 2012 at 8:14pm.
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.
Related Questions
Physics - A rocket is launched from rest. After 8.0 min, it is 160 Km above the ...
physics - How high does a rocket have to go above the Earth’s surface so ...
Physics - PROJECTILE LAUNCHED FROM EARTH'S SURFACE? A projectile is shot ...
physics - The engine of a toy rocket supplies an average acceleration of 38.0 m/...
Physics - A rocket blasts off vertically from rest on the launch pad with an ...
Physics - A rocket starts from rest and moves upward from the surface of the ...
physics - How high does a rocket have to go above Earth's surface until its ...
Calc - A rocket is launched with an initial velocity of zero, and with ...
Calc - A rocket is launched with an initial velocity of zero, and with ...
intro to physics - a toy rocket moving vertically upward passes by a 2.2m- high ...
For Further Reading
