## Poll

13 votes (50%) | |||

10 votes (38.46%) | |||

5 votes (19.23%) | |||

2 votes (7.69%) | |||

8 votes (30.76%) | |||

3 votes (11.53%) | |||

5 votes (19.23%) | |||

4 votes (15.38%) | |||

10 votes (38.46%) | |||

7 votes (26.92%) |

**26 members have voted**

Volume of hemisphere = volume of cone

(2/3) * pi * r^3 = (1/3) * pi * r^2 * h

h=2*r

h=8cm

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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Today I realized that eating ice cream isn't filling the emptiness I feel inside.

But I'm no quitter.

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?

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

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:GialmereIt'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 sec

^{2}and H =138 meters.

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:GialmereIt'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

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 sec^{2}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?

---------------------------------

Here is the pyrotechnician's official mug...

And...

[This is probably my last puzzle until after the new year so, Happy Holidays math gamers!]

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.

Quote:ThatDonGuyActually, 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.