the escape speed of q projectiles on earths surface is11.2 km/a. a body is projected out with twice of this speed. The speed of body far away from the
earth is (Ignore the presence of other bodies)
[NCERT Pg. 202]
(1) 15 km/s
(2) 19.4 km/s
(3) 11.2 km/s
(4) 22.4 km/s
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Answer:
Escape velocity of a projectile from the Earth, v
esc
=11.2 km/s
Projection velocity of the projectile, v
p
=3v
esc
Mass of the projectile =m
Let velocity of the projectile far away from the Earth=v
f
Total energy of the projectile on the Earth =
2
1
mv
p
2
−
2
1
mv
esc
2
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth =
2
1
mv
f
2
From the law of conservation of energy, we have
2
1
mv
p
2
−
2
1
mv
esc
2
=
2
1
mv
f
2
v
f
=
v
p
2
−v
esc
2
=
8
v
esc
=
8
×11.2
=31.68 km/s
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