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At the earth surface a projectile is launched straigh up at the speed of 10 km/s. TO what height will it rise ? Ignore air resistance

Thanks

2007-09-17 20:19:01 · 4 answers · asked by Anonymous in Science & Mathematics Physics

4 answers

Potential energy in an acceleration field is
Ep=mgh
where Ep is potential energy (in Joules)
m is mass (in kg)
h is height (in meters)
When the projectile is moving at 10 km/s it has a kinetic energy of
Ek=(1/2)mv² = (1/2)(1)(10,000)² (assume some mass for the projectile since it will be the same for the potential energy)
So Ex= 5X10^7 and to get the same potential energy, the projectile must be at
5X10^7=(1)(9.8)*(h) or
h=5X10^7/9.8 = 5.102X10^6 meters or 5,102 km.

HTH

Doug

2007-09-17 20:37:05 · answer #1 · answered by doug_donaghue 7 · 0 1

It will rise to an altitude of about 25,501 km above the surface.

You can't use the usual "mgh" value for ΔPE, because that only works if g is unchanging; but in this case the projectile goes so high that the value of g changes significantly.

For the same reason, you can't use "suvat". Again, that only works when the "a" (in this case acceleration due to gravity) is constant.

Epidavros has the right idea, but the wrong formula. The expression "GMm/(R+h)" is the potential energy required to move the projectile from R+h to infinity, not the PE required to move it from the ground to R+h. The correct formula is:

1/2 mv^2 = GMm(1/R – 1/(R+h))

From this formula, combined with the observation that GM/R² = g (acceleration due to gravity at the surface), you can solve for h to get:

h = Rv² / (2gR – v²)

Notice that, if v is small, we have this approximation:

h ~= Rv² / (2gR) = v² / (2g)

which is the same as the "classical" result (constant g).

Also, notice that h approaches infinity as v² approaches the critical value (2gR). This is another way of saying that v = sqrt(2gR) is escape velocity.

2007-09-18 12:39:38 · answer #2 · answered by RickB 7 · 0 0

1/2 mv^2 = GMm/(R+h)

where R = radius Earth and h = height

So:

h = 2GM/v^2 - R

You can only use 1/2mv^2 = mgh if h is very small, because g depends on height.

2007-09-18 03:59:07 · answer #3 · answered by Anonymous · 0 0

its not a question of energy
but suvat

s= ?
u= 10 km/s
or or 10000 m / s

v= 0
a= -9.8


v^2 = u^2 + 2as


wrk it out
i hope it will b fine

tk cr m8

2007-09-18 06:11:38 · answer #4 · answered by Anonymous · 0 1

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