## Poll

16 votes (51.61%) | |||

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10 votes (32.25%) | |||

7 votes (22.58%) |

**31 members have voted**

December 16th, 2020 at 5:59:41 AM
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Here is my solution. I'm omitting the spoiler tags because it's been answered and solved multiple times already.

Let:

t = time since rocket fuel runs out.

r = time rocket fuel lasted.

I'm going to express acceleration in terms of an upward direction. So, the acceleration after the rocket fuels burns out is -9.8.

As a reminder, the integral of acceleration is velocity and the integral of velocity is location. Let's make location relative to the ground.

When the rocket is first launched, we're given that the acceleration is 4.

Taking the integral, the velocity of the rocket after r seconds equals 4r.

Taking the integral of the velocity gives us the location of the rocket after r seconds of 2r^2.

Now let's look at what happens after the rocket fuel burns out.

We're given that acceleration due to gravity is -9.8.

The velocity due to gravity at time t is -9.8t. However, it also has upward velocity of 4r from the rocket.

Let v(t) = velocity at time t

v(t) = -9.8t + 4r

The rocket will achieve a maximum height when v(t) = 0. Let's solve for that.

v(t) = 0 = -9.8t + 4r

4r = 9.8t

t = 40/98 r = 20r/49.

In other words, whatever time the rocket fuel lasted, the rocket will keep traveling up for 20/49 of that time.

We are also given the distance traveled at the maximum altitude achieved is 138.

Let's take the integral of v(t) to get the formula for distance traveled, which we'll call d(t).

d(t) = -4.9t^2 + 4rt + c, were c is a constant of integration.

As we already showed, the rocket traveled 2r^2 by the time the fuel burned out, so that must be the constant of integration. That gives us:

d(t) = -4.9t^2 + 4rt + 2r^2

We know the maximum altitude of 138 was reached at time 20r/49 So let's plug t=20r/49 into the equation to solve for r:

d((20r/49) = -4.9((20r/49)^2 + 4r(20r/49) + 2r^2 = 138

r^2*(-1960/2401 + 80/49 + 2) = 138

r^2 = 49

r = 7

So, the rocket fuel lasted for seven seconds.

We already know the rocket kept going up for 20/49 of that time, which is 140/49 = apx. 2.8571 seconds.

Thus, the time from launch to maximum velocity is 7 + 140/49 = 483/49 = apx. 9.8571 seconds

Let:

t = time since rocket fuel runs out.

r = time rocket fuel lasted.

I'm going to express acceleration in terms of an upward direction. So, the acceleration after the rocket fuels burns out is -9.8.

As a reminder, the integral of acceleration is velocity and the integral of velocity is location. Let's make location relative to the ground.

When the rocket is first launched, we're given that the acceleration is 4.

Taking the integral, the velocity of the rocket after r seconds equals 4r.

Taking the integral of the velocity gives us the location of the rocket after r seconds of 2r^2.

Now let's look at what happens after the rocket fuel burns out.

We're given that acceleration due to gravity is -9.8.

The velocity due to gravity at time t is -9.8t. However, it also has upward velocity of 4r from the rocket.

Let v(t) = velocity at time t

v(t) = -9.8t + 4r

The rocket will achieve a maximum height when v(t) = 0. Let's solve for that.

v(t) = 0 = -9.8t + 4r

4r = 9.8t

t = 40/98 r = 20r/49.

In other words, whatever time the rocket fuel lasted, the rocket will keep traveling up for 20/49 of that time.

We are also given the distance traveled at the maximum altitude achieved is 138.

Let's take the integral of v(t) to get the formula for distance traveled, which we'll call d(t).

d(t) = -4.9t^2 + 4rt + c, were c is a constant of integration.

As we already showed, the rocket traveled 2r^2 by the time the fuel burned out, so that must be the constant of integration. That gives us:

d(t) = -4.9t^2 + 4rt + 2r^2

We know the maximum altitude of 138 was reached at time 20r/49 So let's plug t=20r/49 into the equation to solve for r:

d((20r/49) = -4.9((20r/49)^2 + 4r(20r/49) + 2r^2 = 138

r^2*(-1960/2401 + 80/49 + 2) = 138

r^2 = 49

r = 7

So, the rocket fuel lasted for seven seconds.

We already know the rocket kept going up for 20/49 of that time, which is 140/49 = apx. 2.8571 seconds.

Thus, the time from launch to maximum velocity is 7 + 140/49 = 483/49 = apx. 9.8571 seconds

It's not whether you win or lose; it's whether or not you had a good bet.

December 16th, 2020 at 6:27:14 AM
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Quote:ChesterDogQuote: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.

And since I have nothing better to do at the moment...

Let H be the final altitude, a the rocket's acceleration, and g the absolute value of gravity's acceleration

Let p be the amount of time the rocket was travelling under its own power, and q the amount of time it was subject only to gravity.

After p seconds, the rocket's velocity is ap

After q more seconds, the rocket's velocity is ap - gq; since, in this case, this is zero, ap = gq, so q = (a/g) p.

Break out the basic physics theorems for this:

distance = initial velocity x time + 1/2 x acceleration x (time)^2

Let d be the distance the rocket travelled under its own power; d = 0 + 1/2 a p^2

The distance the rocket travelled while subject only to gravity is (H - d); H - d = ap q - 1/2 g q^2 = a (a/g) p^2 - 1/2 g (a^2/g^2) p^2

H - d = 1/2 a^2/g p^2

H = 1/2 a^2/g p^2 + d = 1/2 a^2/g p^2 + 1/2 a p^2 = 1/2 a p^2 (a/g + 1) = 1/2 p^2 a (a + g) / g

p^2 = 2H g / (a (a + g))

p = sqrt(2H g / (a (a + g))) = sqrt(2H) / sqrt(a (a + g) / g)

The total time = p + q = p (1 + a / g) = p (a + g) / g =sqrt(2H / a) / sqrt((a + g) / g) x (a + g) / g

= sqrt(2H / a) sqrt((a + g) / g) = sqrt(2H (a + g) / ag)

= sqrt(2H (1/a + 1/g)) QED

December 16th, 2020 at 8:42:33 AM
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Today's date 12-16-20 is a Pythagorean triple. The easy math question is, what is the next date that will be a Pythagorean triple?

Poetry website: www.totallydisconnected.com

December 16th, 2020 at 9:05:27 AM
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October 24, 2026 (10^2 + 24^2 = 26^2, which reduces to 5^2 + 12^2 = 13^2

It's not whether you win or lose; it's whether or not you had a good bet.

December 16th, 2020 at 9:17:44 AM
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Mike, I do not concur. There is one before that.

Poetry website: www.totallydisconnected.com

December 16th, 2020 at 9:25:52 AM
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Quote:teliotMike, I do not concur. There is one before that.

July 24, 2025

One back at you: What is the last date in this century that will be a Pythagorean triple?

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

December 16th, 2020 at 12:01:32 PM
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It would be a better world if there were more Pythagorean Triples that could double as dates.Quote:unJonJuly 24, 2025

One back at you: What is the last date in this century that will be a Pythagorean triple?

Poetry website: www.totallydisconnected.com

December 16th, 2020 at 12:06:13 PM
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7/24/25

It’s all about making that GTA

December 17th, 2020 at 11:58:37 AM
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Quote:Ace27/24/25

Isn't 7/25/24 (or 25/7/24 in UK) (25th July 2024) also a possible answer.

December 17th, 2020 at 12:46:35 PM
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Yes. Though a line must be drawn somewhere since there are probably other dates on other calendars (Hebrew, Chinese etc )Quote:charliepatrickQuote:Ace27/24/25Isn't 7/25/24 (or 25/7/24 in UK) (25th July 2024) also a possible answer.

It’s all about making that GTA