A projectile is launched vertically upwards, from earth's surface with speed equal to half of the escape sped upon earth's surface. The maximum height attained is (earth's radius):
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an object, such as a bullet, that is fired from a gun or other weapon
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2.
any object that is thrown as a weapon
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Answer:
Explanation:
If v_{e}v
e
is the escape velocity then
\dfrac{1}{2}v_{e}^2=\dfrac{GMm}{R}
2
1
v
e
2
=
R
GMm
Now the projectile is fired with speed kv_{e}kv
e
.
Then loss in kinetic energy is gain in potential energy
\implies \dfrac{1}{2}m(kv_{e})^2=GMm(\dfrac{1}{R}-\dfrac{1}{h})⟹
2
1
m(kv
e
)
2
=GMm(
R
1
−
h
1
)
\implies k^2(\dfrac{GMm}{R})=GMm(\dfrac{1}{R}-\dfrac{1}{h})⟹k
2
(
R
GMm
)=GMm(
R
1
−
h
1
)
\implies h=\dfrac{R}{1-k^2}⟹h=
1−k
2
R
.
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