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chevy
chevy
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Gialmere
December 14th, 2020 at 8:12:13 AM permalink


Volume of hemisphere = volume of cone
(2/3) * pi * r^3 = (1/3) * pi * r^2 * h
h=2*r

h=8cm
Gialmere
Gialmere
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December 14th, 2020 at 3:13:33 PM permalink
Quote: rsactuary



h=8

using volume of a hemisphere = 4/6 x pi x r^3

and volume of a cone = pi x r^2 X h/3


Quote: chevy



Volume of hemisphere = volume of cone
(2/3) * pi * r^3 = (1/3) * pi * r^2 * h
h=2*r

h=8cm


Correct!
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Gialmere
Gialmere
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December 15th, 2020 at 8:07:23 AM permalink
It's toughie Tuesday. Time for rocket science...



You are a pyrotechnician in charge of the nightly firework display at an amusement park. You've received some new style rockets from Europe and are testing one in order to time it to your show's music soundtrack.

The firework rocket is fired vertically upwards with a constant acceleration of 4 ms^-2 until the chemical fuel expires. Its ascent is then slowed by gravity until it reaches a maximum height of 138 meters where it detonates.

Assuming no air resistance and taking g=9.8 ms^-2, how long does it take the rocket to reach its maximum height?
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Wizard
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Wizard
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Thanks for this post from:
Gialmere
December 15th, 2020 at 9:46:26 AM permalink
Quote: Gialmere

Assuming no air resistance and taking g=9.8 ms^-2, how long does it take the rocket to reach its maximum height?



Apx. 9.85 seconds.

Not that you asked, but the rocket fuel burns out in seven seconds.

I'll provide a solution (hopefully) if my answer is right
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ChesterDog
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Gialmere
December 15th, 2020 at 10:37:13 AM permalink
Quote: Gialmere

It's toughie Tuesday. Time for rocket science... ...
You are a pyrotechnician in charge of the nightly firework display at an amusement park. You've received some new style rockets from Europe and are testing one in order to time it to your show's music soundtrack.

The firework rocket is fired vertically upwards with a constant acceleration of 4 ms^-2 until the chemical fuel expires. Its ascent is then slowed by gravity until it reaches a maximum height of 138 meters where it detonates.

Assuming no air resistance and taking g=9.8 ms^-2, how long does it take the rocket to reach its maximum height?




I agree with the Wizard's answer.

Expressed algebraically, I get a total time of [ 2H ( 1/a + 1/g ) ]1/2, where a = 4 meters per sec2 and H =138 meters.

ThatDonGuy
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Gialmere
December 15th, 2020 at 3:02:50 PM permalink

Let a be the time the fuel burns
At time a, the distance = 1/2 x 4 x a^2 = 2 a^2
and the velocity = 4 a

Let b be the time from when the fuel runs out to when the rocket explodes
At time b, the distance = 4 ab + 1/2 x (-9.8) x b^2 = 4 ab - 4.9 b^2
and the velocity = 4 a - 9.8 b = 0
a = 2.45 b

The total distance = 2 a^2 + 4 ab - 4.9 b^2 = 138
2 (2.45 b)^2 + 9.8 b^2 - 4.9 b^2 = 138
(4.9 x 2.45 + 4 x 2.45 - 2 x 2.45) b^2 = 138
b^2 = 138 / (6.9 x 2.45)
b = sqrt(138 / (6.9 x 2.45))
The total distance = a + b = 3.45 sqrt(138 / (6.9 x 2.45)) = 69 / 7 m

ssho88
ssho88
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Gialmere
December 15th, 2020 at 4:01:11 PM permalink
Quote: Gialmere

It's toughie Tuesday. Time for rocket science...



You are a pyrotechnician in charge of the nightly firework display at an amusement park. You've received some new style rockets from Europe and are testing one in order to time it to your show's music soundtrack.

The firework rocket is fired vertically upwards with a constant acceleration of 4 ms^-2 until the chemical fuel expires. Its ascent is then slowed by gravity until it reaches a maximum height of 138 meters where it detonates.

Assuming no air resistance and taking g=9.8 ms^-2, how long does it take the rocket to reach its maximum height?





g=9.8, a = 4, H = 138

t1 =( 2gH/(ag+a^2))^0.5 = 7

t2 = (2aH/(g^2+ag))^0.5 = (1104/135.24)^0.5 = 2.857

Total time = 9.857 sec

S1 = 1/2a(t1)^2 = 98

S2 = u(t2) - 1/2g(t2)^2 = 4*7*(1104/135.24)^0.5 - 4.9*(1104/135.24) = 40

Gialmere
Gialmere
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Thanks for this post from:
unJon
December 15th, 2020 at 5:05:05 PM permalink
Quote: Wizard


Apx. 9.85 seconds.

Not that you asked, but the rocket fuel burns out in seven seconds.

I'll provide a solution (hopefully) if my answer is right


Quote: ChesterDog



I agree with the Wizard's answer.

Expressed algebraically, I get a total time of [ 2H ( 1/a + 1/g ) ]1/2, where a = 4 meters per sec2 and H =138 meters.


Quote: ThatDonGuy


Let a be the time the fuel burns
At time a, the distance = 1/2 x 4 x a^2 = 2 a^2
and the velocity = 4 a

Let b be the time from when the fuel runs out to when the rocket explodes
At time b, the distance = 4 ab + 1/2 x (-9.8) x b^2 = 4 ab - 4.9 b^2
and the velocity = 4 a - 9.8 b = 0
a = 2.45 b

The total distance = 2 a^2 + 4 ab - 4.9 b^2 = 138
2 (2.45 b)^2 + 9.8 b^2 - 4.9 b^2 = 138
(4.9 x 2.45 + 4 x 2.45 - 2 x 2.45) b^2 = 138
b^2 = 138 / (6.9 x 2.45)
b = sqrt(138 / (6.9 x 2.45))
The total distance = a + b = 3.45 sqrt(138 / (6.9 x 2.45)) = 69 / 7 m

Quote: ssho88




g=9.8, a = 4, H = 138

t1 =( 2gH/(ag+a^2))^0.5 = 7

t2 = (2aH/(g^2+ag))^0.5 = (1104/135.24)^0.5 = 2.857

Total time = 9.857 sec

S1 = 1/2a(t1)^2 = 98

S2 = u(t2) - 1/2g(t2)^2 = 4*7*(1104/135.24)^0.5 - 4.9*(1104/135.24) = 40


Correct!

Very good.

Wow! We have quite a few rocket scientists here. How are you guys at brain surgery?
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ThatDonGuy
ThatDonGuy
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December 15th, 2020 at 7:17:10 PM permalink
Quote: Gialmere


Wow! We have quite a few rocket scientists here.


Actually, real rocket scientists would have pointed out that gravity is not constant, but varies as the inverse square of the distance to the center of gravity, and you also have to take into account the decrease in the rocket's mass as the fuel is used up.
ssho88
ssho88
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December 15th, 2020 at 8:33:07 PM permalink
Quote: ThatDonGuy

Actually, real rocket scientists would have pointed out that gravity is not constant, but varies as the inverse square of the distance to the center of gravity, and you also have to take into account the decrease in the rocket's mass as the fuel is used up.



Air resistance/surface resistance should decrease when rocket go higher, will surface resistance decrease when surface temperature increase ? The Coriolis Effect ?

For knowledge purposes only.
Last edited by: ssho88 on Dec 15, 2020

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