# Posts by TchrWill

Total # Posts: 701

**Physics**

The equations of accelerated motion, derived elsewhere, also apply to falling (or rising) bodies with the exception that the term "a" for acceleration is replaced by the term "g” for the acceleration due to gravity). This results in ....Vf = Vo + gt (the ...

*April 11, 2013*

**Geometry**

The area of a spherical triangle bounded by arcs of three great circles is defined by A = Pi(r^2)E/180 where r = the spherical radius and E = the spherical excess, defined by E = (A+B+C)-180, A, B and C being the three spherical angles of the sphere.

*March 5, 2013*

**wireless communications**

The velocity required to maintain a circular orbit around the Earth may be computed from the following: Vc = sqrt(µ/r) where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of the earth, ~...

*December 2, 2012*

**Physics**

Using the radius of the Earth and Moon as 3963 miles and 1080 miles, respectively, and the distance between them as 239,000 miles, the surface to surface distance becomes 233,960 miles. From Vf^2 = Vo^2 - 2gs, 0 = Vo^2 - 2(32.2)233,960 or Vo = 282,053 fps = 192,309 mph. The ...

*November 15, 2012*

**Physics**

The velocity required to maintain a circular orbit around the Earth may be computed from the following: Vc = sqrt(µ/r) where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of the earth, ~...

*November 8, 2012*

**physics**

Contrary to popular belief, a spacecraft does not have to reach full escape velocity in order to reach the Moon. The full escape velocity from a 200 mile high circular orbit is 24,400 mph. Assuming the trip starts from a 200 mile high circular orbit, the minimum injection ...

*October 24, 2012*

**Physics**

There is a clear distinction between a geosynchronous orbit and a geostationary orbit. The early recognition of a geostationary orbit was made by the Russian Konstantin Tsiolkovsky early this century. Others referred to the unique orbit in writings about space travel, space ...

*October 16, 2012*

**math**

The formula that will clear up your problem is ......Vc = sqrt(µ/r) where Vc = the velocity required to keep a body in a circular orbit, in feet/sec., r = the orbital radius in feet and µ = the Earth's gravitational constant or 1.407974x10^16 ft^3/sec^2. The ...

*August 4, 2012*

**Algebra**

2X + (X - 4) = 32 - X

*July 29, 2012*

**Physics**

Can you explain wheightlessness. Do you actually weigh nothing in space or just a percentage of your Earth weight? How much would a 100 lb person weigh on each of the other planets? >> An astronaut, circling the earth in the Space Shuttle, senses that his apparent weight...

*July 8, 2012*

**Physics**

There is a clear distinction between a geosynchronous orbit and a geostationary orbit. The early recognition of a geostationary orbit was made by the Russian Konstantin Tsiolkovsky early this century. Others referred to the unique orbit in writings about space travel, space ...

*June 29, 2012*

**physics**

Time out = 6/2.5 = 2.4 hours Time back = 6/4 = 1.5 hours Average speed for the round trip = (6 + 6/(2.4 + 1.5) = 3.077 kmph

*May 25, 2012*

**physics**

Since the velocity of the train is constant at 54kmph, the average velocity is 54kmph. Proof: D1 = 54(.5) = 27km. D1 = 54(.6666666) = 36 Vav = (27 + 36)/1.16666 = 54kmph.

*May 25, 2012*

**geometry**

* The altitude to the hypotenuse of a right triangle creates two similar triangles, each similar to the original right triangle and to each other. * The altitude to the hypotenuse of a right triangle is the geometric mean between the segments of the hypotenuse created by the ...

*May 23, 2012*

**Math**

The numbers 4; 6; 13; 27; 50; 84 do not form an arithmetic progression as the differences between succsessive terms are not constant. If you take the successive differences of the terms given, n.......1....2....3....4....5....6... N.......4....6...13...27...50...84... 1st Diff...

*May 11, 2012*

**inscribed angles**

Clearly, I did not understand you. But after reviewing the scenario, the answer is clearly d = sqrt(17000^2 + 4000^2)-4000 = 13,464

*May 10, 2012*

**inscribed angles**

If I understand you correctly the distance you seek is d = sqrt(17000^2 - 4000^2) - 4000

*May 10, 2012*

**Physics**

Given the three sides (the largest must be less than the sum of the other two) Find any angle from cosA = (b^2 + c^2 - a^2)/2bc. The remaining angles can be derived from the Law of Sines. Alternatively, find any angle, A for instance, using tan(A/2) = r/(s - a) where s = (a + ...

*May 6, 2012*

**physics**

Thye following is more than you asked for but I hope you find it interesting. I think you would be surprised to know just how many people have no idea that the Earth moves at all, thinking that the Sun revolves around the Earth. The motion of our Earth through space is really ...

*April 17, 2012*

**physics**

Moment of inertia M(i) = 2WR^2/5 about any axis.

*April 17, 2012*

**Geometry**

3.14(11^2)/6 - 2(11)5.5/2 = 63.355 - 60.5 = 2.855 sqcm Therefore, D.

*April 15, 2012*

**Algebra**

A movie theater manager wants to know how many adults and how many children pay admission to a particular movie. The theater charges $10.00 for adult tickets and $5.00 for child tickets. At a showing where 210 tickets were sold the theater collected $1,355.00. Write a system ...

*March 25, 2012*

**Algebra**

Let A = the number of adult tickets C = the number of children tickets Then, A + C = 210 and 10A + 5C = 1355 Can you take it from here?

*March 25, 2012*

**algebra**

Continuous Compound Interest Formula where FV = Pe^(rt) P = principal amount (initial investment) r = annual interest rate (as a decimal) t = number of interest bearing years FV = amount after time t 110,682 = 45,000e^(30r) from which r = .03 or an annual rate of 3%.

*March 18, 2012*

**trig**

Part of my previous reply was lost in the posting process. The orbital radius is 3960 + 2.1(104) = 4178.4 miles or 22,061,952 feet. The alleged time to complete one orbit is 10.5(24)3600 = 907,200 seconds making the derived orbital velocity Vc = 152.7fps. Unfortunately, the ...

*March 17, 2012*

**trig**

/r] sqrt(sqrt[(1.407974x10^16)/(3960+218.4)5280] = 25,262 feet per second. The orbital period is T = 2(3.14)sqrt[22,061,952^3/1.407974x10^16] = 5487 seconds or 91.45 minutes.

*March 17, 2012*

**physics**

As we normally think of it, the static value of gravity on, or above the surface of a spherical body is directly proportional to the mass of the body and inversely proportional to the square of the distance from the center of the body and is defined by the expression g = GM/r^...

*March 13, 2012*

**Physics**

An asteriod is found a distance from the sun equal to 32 units times the earth's distance. What will be the length of time in years required for this asteroid to make one revolution around the sun? Rast = 32 x 92,960,242 = 2,974,727,744 miles GMsun = gravitational constant...

*March 10, 2012*

**physics**

The acceleration due to gravity derives from g = µ/R^2 where G = the acceleration due to gravity µ = the earths gravitational constant = GM G = the universal gravitational constant = 6.67259x10^-11 M = the mass of the earth = 5.97424x10^24 R = the mean surface ...

*December 15, 2011*

**physics**

When orbiting at 5.8km/s: From Vc = sqrt(µ/r) where Vc = the velocity of an orbiting body, µ = the gravitational constant of the earth and r the radius of the circular orbit,with µ = GM, G = the universal gravitational constant and M = the mass of the central...

*December 11, 2011*

**physics**

When orbiting at 6.3km/s: From Vc = sqrt(µ/r) where Vc = the velocity of an orbiting body, µ = the gravitational constant of the earth and r the radius of the circular orbit,with µ = GM, G = the universal gravitational constant and M = the mass of the central...

*December 10, 2011*

**physics**

g = µ/r^2 gravity on the surface of Mars µ = the gravitational constant of Mars = GM where G = the universal gravitational constant and M = the planet mass r = the radius of Mars = .53(6378)1000 = 3,380,340m G = 6.67259x10^-11 M = .11(5.97424x10^24) = 6.571664x10^...

*December 9, 2011*

**Physics**

For Vex = velocity of the exhaust gases g = acceleration due to gravity w = the flow rate of the exhaust gases F = the thrust exerted on the rocket Vex/g = F/w or F = wVex/g

*December 9, 2011*

**Math**

37 + 10A + B = 59 10A + B = 22 10A + 2 = 22 10A = 20 making A = 2 or 37 + 22 = 59

*December 9, 2011*

**math**

x(x + 1) = x + (x + 1) + 11 Your move.

*December 8, 2011*

**algebra**

What is the monthly deposit required to accumulate to a fund of $1,000,000 over a period of 40 years with deposits starting at the end of the first month and bearing an interest rate of 8% compounded monthly? S(n) = $1,000,000] i = .08/12 = .00666... n = 40(12) = 48 R = S(n)(i...

*December 8, 2011*

**algebra**

Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area? Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x) = Px - 2x...

*December 8, 2011*

**Physics**

The following might be of some help to you. 1/23/02 What is the final burnout velocity of the Space Shuttle? The final burnout velocity of the Space Shuttle, or any rocket for that matter, is a function of the altitude at burnout and the orbit that the rocket is being injected...

*December 7, 2011*

**science**

Assuming you are seeking the altitude where gravity = g = 6.5m/s^2: As we normally think of it, the static value of gravity on, or above the surface of a spherical body is directly proportional to the mass of the body and inversely proportional to the square of the distance ...

*December 6, 2011*

**math**

A convex polyhedron is defined as a solid with flat faces and straight edges so configured as to have every edge joining two vertices and being common to two faces. There are many convex polyhedra, only five of which are considered regular polyhedra. Regular polyhedra satisfy ...

*December 6, 2011*

**math**

Lets look at a typical problem with real numbers. If there are 15 people in a room and each person shakes hands with every other person in the room only once, how many handshakes will take place? There are two ways of looking at this type of problem. The first involves ...

*December 5, 2011*

**Physics**

Vc = sqrt(µ/r) V(A)^2 = µ/r(A) V(B)^2 = µ/r(B) = µ/2r(A) V(B)^2/(2/V(A)^2 = [µ/2r(A)]/[µr(A)] .................= 2 V(B)/V(A) = sqrt2

*December 5, 2011*

**algebra**

x + y = 172 110x + 235y = 29,420. Can you take it from here?

*December 1, 2011*

**Algebra**

Considering all rectangles with the same perimeter, the square encloses the greatest area. Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x...

*December 1, 2011*

**Physics**

From Vc = sqrt(µ/r) where Vc = the velocity of an orbiting body, µ = the gravitational constant of the earth and r the radius of the circular orbit,with µ = GM, G = the universal gravitational constant and M = the mass of the central body, the earth in this ...

*December 1, 2011*

**Math**

The formula for calculating a monthly loan payment is R = Pi/[1 - 1/(1+i)^n] where R = the periodic payment, P = the principal, or debt to be paid off, n = the number of payment periods over which the payments will take place, and i = the periodic interest rate in decimal form...

*November 28, 2011*

**math**

The present value of an ordinary annuity is the sum of the present values of the future periodic payments at the point in time one period before the first payment. What is the amount that must be paid (Present Value) for an annuity with a periodic payment of R dollars to be ...

*November 28, 2011*

**physics**

The velocity required to maintain a circular orbit around the Earth may be computed from the following: Vc = sqrt(µ/r) where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of the earth...

*November 28, 2011*

**Algebra**

Letting x, y, and z equal the numbers of .02, .08 and .14 cent stamps. 1-x = 2y + 5 2-z = 3y 3-.02x + .o8y + .14z = 2.26 4-Times 100 yields 2x + 8y + 14z = 226 5-Substituting (1) and (2) into (4) yields ..2(y + 5) + 8y + 14(3y) 6-Can you take it from here?

*November 22, 2011*

**Algebra II**

The Law of Universal Gravitation states that each particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Expressed ...

*November 21, 2011*

**gravity**

Addendum to my earlier response. A reduction in velocity of only 1.5m/sec. will drop the space shuttle into an ever so slightly elliptical orbit with an apogee of 6828km, a perigee of 6825.6km. and a period of 5611.86 sec. or 93.53min., allowing the shuttle to catch up with ...

*November 19, 2011*

**gravity**

Assume, for discussion purposes, a satellite and the shuttle are in the same 91 minute circular orbit, the satellite 2 minutes ahead of the space shuttle. The shuttle fires its on-orbit thrusters to reduce its orbital velocity. In doing so, the space shuttle drops into an ...

*November 19, 2011*

**physics**

The equations that apply to rising bodies are .....Vf = Vo - gt (the term “g” for acceleration due to gravity is assumed constant on, or near, the surface of the Earth) .....d = Vo(t) - g(t^2)/2 .................2 .....Vf^2 = Vo^2 - 2gd From Vf^2 = Vo^2 - 2gd .....26...

*November 18, 2011*

**Physics**

L = the left and and The summation of moments about the left end support is 1.4(55) + 3.8(70) - 5R. Solve for R, the right end support load. The summation of vertical forces is L - 55 - 70 + R. Solve for L.

*November 17, 2011*

**math**

There is an old rule of thumb that states that the time required for an investment to double is equal to 72 divided by the interest rate in percent. Example, for $10,000 to double to $20,000 at 6% interest, compounded annualy, it would have to remain on deposit for 72/6 = 12 ...

*November 17, 2011*

**math 140**

This is an ordinary annuity where R dollars is deposited in a bank at the end of each month and earning interest compounded monthly. S(n) = R[(1+i)^n - 1]/i where R = the monthly deposit, S(n) = the ultimate accumulation, n = the number of periods the deposits are made and i...

*November 15, 2011*

**Physics**

Since there is no atmosphere in space, the typical control surfaces such as the aileron, rudder, or elevator, used on an airplane would serve no purpose. All spacecraft use what is called an Attitude Control System(ACS) or a Reaction Control System(RCS) to control the attitude...

*November 15, 2011*

**math**

Ignoring the corner posts, there are 14 posts along each long side, 9 posts along each short side and the 4 corner posts for a total of 2(14) + 2(9) + 4 = 50

*November 15, 2011*

**physics**

A rocket is fired in deep space, where gravity is negligible. In the first second it ejects 1/160 of its mass as exhaust gas and has an acceleration of 15.1 m/s^2.What is the speed v_gas of the exhaust gas relative to the rocket? By means of the rocket equation, deltaV = cln(...

*November 14, 2011*

**math**

Having spent $25, LEa realizes that she still has 60% of her original sum. Therefore, $25 must represent 40% or her original sum making the original sum 25/.40 = $62.50. $62.50 - $25.00 = $37.50. $37.50/$62.50 = .6 or 60%. $3.00/2.6 = $1.15 per liter

*November 14, 2011*

**algebra**

From P R[1 - (1+i)^(-n)]/i where P = present value R = the periodic payment i = the decimal interest per compounding period = I/100 assuming interest compounded anually and n = the number of interest bearing periods. Therefore, P = 7300[1 - 1.0070833^(-16)]/.0070833 p = $110,056

*November 14, 2011*

**Astronomy**

The eccentricity of Mar's orbit is .0934. The semi-major axis is 141,643,675 miles. Therefore, the perihelion distance is r(p) = 141,643,675(1-.0934) = 128,414,418. and r(a) = 141,643,675(1+.0934) = 154,871,931. The mean radius of the earth's orbit is r = 92,960,242 ...

*November 14, 2011*

**Physics**

From Vf^2 = Vo^2 + 2gh Vf^2 = 0 + 2(9.8)250,000 making Vf = 2213m/s or From h = Vo(t) = g(t^2)/2 250,000 = 9.8t^2/2 making t = 225.87sec. and Vf = Vo + gt Vf = 0 + 9.8(225.87) = 2213m/s

*November 13, 2011*

**Math**

How about 43^5.

*November 11, 2011*

**math**

P = 9n n = 1 -2 -3 -4 -5 -6 -7- 8- 9 -10 P = 9-18-27-36-45-54-63-72-81-100 Notice anything?

*November 9, 2011*

**Math**

Might you have noticed the new HD screen ratio comparison to its predecessor? The old screen ratio was 4:3 while the new one is 16:9. That's 4^2/3^2. The new screen ratio actually results in an 11% reduction in actual screen area.

*November 9, 2011*

**Math**

Compound Interest With compound interest, the interest due and paid at the end of the interest compounding period is added to the initial starting principal to form a new principal, and this new principal becomes the amount on which the interest for the next interest period is...

*November 9, 2011*

**Math, Finance**

You have the right formula typically shown in the form or Ri = Sn/[(1+i)^n - 1] Sn = 120,000 n = 15x2 = 30 i = .068/2 = .034 Make the keys dance.

*November 9, 2011*

**Physics**

Escape velocity derives from Ve = sqrt[2µ/r] where Ve = escape velocity in ft/sec., µ = the gravitational constant of the earth and r = the surface radius. µ can also be stated as GM where G = the gravitational and M = the mass of the planet. Therefore, Ve ...

*November 9, 2011*

**Physics**

Vp/Va = r(a)/r(p)

*November 9, 2011*

**PHYSIC**

Since 8000 - 1630 = 6370, the radius of the earth in km, I believe I am on safe ground assuming that you mean "740km" above the earth in your first question. The velocity required to maintain a circular orbit derives from Vc = sqrt(µ/r) where Vc = the velocity ...

*November 8, 2011*

**Math**

If she had 6 left after giving 3/5 to her children, she had 6/(2/5) = 15 to start. If she had 15 left after giving 1/4 to her neighbors, she had 15/(3/4) = 20 to start. If she had 20 left after giving 1/2 to her sister, she had 20/(1/2) = 40 to start out with in the first ...

*November 5, 2011*

**advanced math**

Alternatively: Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area? Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P...

*November 3, 2011*

**advanced math**

Let the sides parallel to the given boundary be x and the sides perpendicular to the boundary be y. Then, 2x + 3y = 168 and A = 2xy. With y = (168 - 2x)/3 A = (336x - 4x^2)/3 dA/dx = 112 - 8x/3 = 0 336 = 8x making x = 84 and y = 28 The area A = 2352m^2.

*November 3, 2011*

**Physics**

The Law of Universal Gravitation states that each particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Expressed ...

*November 3, 2011*

**geometry**

Considering all rectangles with a given perimeter, which one encloses the largest area? The traditional calculus approach would be as follows. Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x)/2 = Px/...

*November 3, 2011*

**physics180**

A satellite is in circular orbit at a height R above earth's surface. a)find orbital period. b)what height is required for a circular orbit with a period double that found in part (a)? Allow me to modify your terms. a) The orbit radius R = Re + h where Re = the earth's...

*November 2, 2011*

**physics**

The average speed for the entire trip is the total distance divided by the total time. Therefore, t1 = 5/6.2 = .806 hr. t2 = 5/V Total distance traveled is 5 + 5 = 10 miles. Total time is t1 + t2 = .806 + 5/V Average driving speed is therefore Vavg = 10/(.806 + 5/V) = 10.6 ...

*November 1, 2011*

**algebra**

There are 132 more silver dollars than 50-cent pieces or D = H + 132. The ratio of silver dollars to 50-cent pieces is 5:2 or D/H = 5/2. How many of each type of coins are in Mr. Hundley's collection? D = 5H/2 Therefore, 5H/2 = H + 132 5H = 2H + 264 3H = 264 H = 88 D = 88...

*November 1, 2011*

**Algebra 2**

Using V = D/T 60D/45 = T - 1 60D/40 = T + 1 Solving for T and equating yields 60(40D - 45D)/1800 = 2 or 5D = 60 making D = 12 miles In order to arrive in 17 minutes, 60(12)/V = 17 yielding = 42.35mph.

*November 1, 2011*

**physics**

The orbital period of a satellite derives from T = 2(Pi)sqrt(r^3/µ) where T = the period in seconds Pi = 3.14 r = the orbital radius in feet and µ = the earth's gravitational constant = 1.407974x10^16 ft^3/sec^2. 139(60)= 2(3.14)sqrt(r^3/1.407974x10^16) Solving...

*October 31, 2011*

**Physics**

Given the distance and angle of elevation, the velocity with which the ball left the club face derives from d = V^2(sin(µ))/g which then allows the maximum height reached to be found from h = [V^2(sin^2(µ))]/2g

*October 24, 2011*

**physics**

How do you determine the altitude at which a satellite must fly in order to complete one orbit in the same time period that it takes the earth to make one complete rotation? The force exerted by the earth on the satellite derives from...

*October 23, 2011*

**Physics**

The period of an orbiting body derives from ....T = 2(Pi)sqrt(r^3/µ) in seconds where T = the period in sec, Pi = 3.1416, r = the orbital radius about the Sun in meters and µ = the planet's gravitational constant = Gm where G = the gravitational constant of ...

*October 21, 2011*

**math**

For a given perimeter, the rectangle that encloses the maximum area using one side as a building, river, fence, etc. or an adjacent section of fence, is in the ratio of 2/1. Assuming each of the 16 grazing areas is in the length to width ratio of 2/1: Let the length of the 16 ...

*October 19, 2011*

**PHYSÝCS**

I must admit I am not clear on what you mean by "mean radius" of the earth; average, arithmetic, geometric, harmonic, etc. I am making my own interpretation to be the radius that divides the spherical earth into two equal volumes, ignoring the equatorial bulge. Using...

*October 17, 2011*

**math**

Vf = Vo(t) - 32(t) Vf = 88 - 32t or t = 2.75sec. Height reached from 25 ft. height h = 88(t) - 32(t^2)/2 h = 88(2.75) - 16(2.75)^2 = 121 ft. It takes the same 2.75 sec. to return to the original launch height of 25 feet and an increment more to fall the 25 feet further to the ...

*October 16, 2011*

**algebra**

Let the corner pieces be x by x. Then, x(16 - 2x)(9 - 2x) = 120 144x - 50x^2 + 4x^3 = 120 4x^3 - 50x^3 + 144x - 120 = 0 2x^3 - 25x^2 + 72x - 60 = 0 First derivative = 6x^2 - 50x + 72 = 0 3x^2 - 25x + 36 = 0 x = [25+/-sqrt(25^2 - 4(3)36)]/6 x = 1.851

*October 16, 2011*

**Physics 1 - Mechanics**

Being in a 261 mile high orbit, your orbital velocity is 25,125 fps. The orbital period is 92.95 minutes. Assuming that the 21km (68,901 ft.) distance is along the circumference of the orbit, your friend is only 2.75 sec. ahead of you. Clearly, you need only slow down a minute...

*October 14, 2011*

**physics**

The weight is 2.0x106 x 2.205 = 4,410,000 lbs. (actually 4,458,000 with a 25,000 lb. payload) Liftoff thrust is 35x10^6N .102 x 2.205 = 7,871,850 lbs. (actually 7,245,000 lbs.) Net liftoff acceleration a = 7,871,850(32.2)/4,410,000 - 32.2 = 25.27fps^2. Velocity attained after ...

*October 14, 2011*

**Math WORD PROBLEM**

The actual velocity of a satellite orbiting at 860 miles altitude is 24,113 fps. The period derives from T = 2(Pi)sqrt(r^3/µ) = 113.3 minutes. The actual velocity of a satellite orbiting at 1700 miles altitude is 21,705 fps. The period derives from T = 2(Pi)sqrt(r^3/&...

*October 12, 2011*

**Algebra**

1--.05x-.6(x+2)=.08 2--Multiplying by 100 yields ...5x - 60x - 120 = 8 3--Collecting terms, -55x = 128 4--Therefore x = -128/55 5--Checking: ...05(-128/55) - .6(-128/55 - 1.2 = .08 ...Multiplying through by 55 yields ....05(-128) - .6(-128) - 66 = 4.4 6----6.4........+.76.8...

*October 12, 2011*

**physics**

Excuse the change in units. 60 year old habits are hard to break. You left us without a clue as to the weight of the rocket. Propellant fractions of rockets typically range from .85 to .90, i.e., the weight of propellant divided by the fully loaded rocket stage. For this ...

*October 11, 2011*

**Physics**

d = (24.5)^2(sin(2*35º)/9.8 = h = (24.5)^2(sin^2(35º)/2(9.8) =

*October 10, 2011*

**Physics**

g = .379(9.8) = 3.71m/s^2 Up time = down time = 3.1sec. each. Vf = 0 = Vo - 3.71(3.1) making Vo = 11.5m/s h = 11.5(3.1) - 3.71(3.1)^2/2 = 17.82m (a) 17.82m (b) Vo = 11.5m/s.

*October 10, 2011*

**algebra**

(30t)^2 + (40t)^2 = 400^2

*October 10, 2011*

**algebra**

The Golden Ratio = 1.618. Therefore, the length would be 1.618(14.72) = 23.817.

*October 9, 2011*

**trig**

There is insufficient information to determine the height after traveling 6000 ft. horizontally. It will definitely be on an angle less than 18º from the launch point.

*October 9, 2011*

**Algebra**

1--70t1 + 20t2 = 260 2--t1 + t2 = 9 or t1 = 9 - t2 3--Substitute (2) into (1) and solve for t2 4--Distance traveled at 20mph = 20t2

*October 9, 2011*

**trig**

(H/tan8º)^2 + (H/tan11º)^2 = 1500^2

*October 9, 2011*