View Full Version : vertical motion: h = -16t^2 + vt + s

kaquinlan

06-06-2006, 02:04 PM

I am trying to do a vertical motion problem and it keeps coming up a ridiculously negative number!

I know the formula is:

. . .h = -16t^2 + vt + s, where

. . ."h" is the height (in feet) at a given time

. . ."s" is the initial height (in feet)

. . ."t" is the time in motion (in seconds)

. . ."v" is the initial velocity (in feet per second)

If the initial height is 8 ft and the initial velocity is 95 feet/sec, how high will the object be in 4 minutes?

I used:

. . .h = -16(240)^2 + 95(240) + 8

. . .h = -16(57,600) + 22,800 + 8

. . .h = -921,600 + 22,800 + 8

. . .h = -898,792

Obviously wrong! What am I doing wrong?

stapel

06-06-2006, 02:13 PM

It really says "after four minutes", not "four seconds"...?

Because if it does, then (ignoring the fact that the object would have hit the ground and either bounced or stopped), the immense negative answer is mathematically correct.

Thank you.

Eliz.

skeeter

06-06-2006, 02:18 PM

uhh ... the object will only be in the air for a little over 6 seconds, after which it will hit the ground.

:shock:

so, after 4 minutes, the object will have been at rest on the ground for about 3 minutes and 54 seconds ... its height will be 0.

are you sure the problem didn't say 4 seconds :?:

kaquinlan

06-06-2006, 02:25 PM

This still makes no sense to me and the object is a rocket? It says the rocket takes flight from a platform that is 8ft high. It has an initial velocity of 95 ft/sec. How high will the rocket be after 4 minutes of flight? Would it still be negative?

skeeter

06-06-2006, 02:49 PM

Wait a minute ... now it's a rocket? A rocket is not a projectile until it is acted upon solely by the force of gravity, i.e. when it has run out of propellant. ... did you leave out any information? I recommend that you cite the original problem in its entirety.

kaquinlan

06-06-2006, 03:42 PM

It says:

The formula for the vertical motion of an object is:

h = -16t^2 + vt + s, where:

. . ."h" is the height (in feet) at a given time

. . ."s" is the initial height (in feet)

. . ."t" is the time in motion (in seconds)

. . ."v" is the initial velocity (in feet per second)

In the model, the coefficient of t^2 is one half of the acceleration due to gravity. On the surface of the earth this acceleration is approx 32 ft per second.

A rocket takes flight from a platform that is 8ft high. It has an initial velocity of 95 ft per second. How high will the rocket be after 4 minutes of flight?

stapel

06-06-2006, 03:45 PM

And the exercise says to use the formula for projectile motion (the motion of something not under acceleration after launch) for a rocket (an object that is under acceleration after launch)?

If so, then you have already received explanations of the answer. If not, then please reply with a corrected version of the instructions and/or the formula you are actually expected to use.

Thank you.

Eliz.

kaquinlan

06-06-2006, 04:12 PM

Then the next problem would be negative too??

"A rocket takes off from a launcher that is 3 feet high. It has an initial velocity of 60 feet per second. After 5 minutes of flight how high will the rocket be?"

Sounds like the same thing?

stapel

06-06-2006, 05:22 PM

As has been explained, either the equation is wrong (since it doesn't fit a rocket), or else the definition is wrong (it isn't actually a rocket), or else the time is wrong (it's seconds, not minutes), or else the answer must, of necessity, be a nonsensical negative number.

Unless you can provide clarification as to which of the above is correct, there is no way for us to give you any different explanation and answer than has already been given to you.

Eliz.

kaquinlan

06-07-2006, 08:18 AM

The formula is correct and the numbers are correct. This would be a model rocket - we are building them in school

stapel

06-07-2006, 11:45 AM

The formula is correct and the numbers are correct. This would be a model rocket - we are building them in school

If your instructor really thinks that the formula for non-propelled vertical flight (simple one-dimensional projectile motion) actually applies to something with a motor flying at an angle, then somebody in authority needs to have a serious talk with the instructor.

Eliz.

ScienceWorld: The Rocket Equation (http://scienceworld.wolfram.com/physics/RocketEquation.html)

Wikipedia: Tsiolkovsky rocket equation (http://en.wikipedia.org/wiki/Rocket_equation)

Answers.com: Tsiolkovsky rocket equation (http://www.answers.com/topic/tsiolkovsky-rocket-equation)

Equations for Model-Rocketeers (http://my.execpc.com/~culp/rockets/rckt_eqn.html)

Rocket and Space Technology: Rocket Propulsion (http://www.braeunig.us/space/)

Ask Dr. Math: Simplified Rocket Kinematics (http://mathforum.org/library/drmath/view/56364.html)

Vertical Projectile Motion (http://www.purplemath.com/modules/quadprob.htm)

TchrWill

06-07-2006, 02:37 PM

I am trying to do a vertical motion problem and it keeps coming up a ridiculously negative number!

I know the formula is:

. . .h = -16t^2 + vt + s, where

. . ."h" is the height (in feet) at a given time

. . ."s" is the initial height (in feet)

. . ."t" is the time in motion (in seconds)

. . ."v" is the initial velocity (in feet per second)

If the initial height is 8 ft and the initial velocity is 95 feet/sec, how high will the object be in 4 minutes?

I must assume that you meant 4 seconds and not 4 minutes.

I must also assume that you mean the projectile (rocket) leaves the 8 foot high stand with an instantaneous velocity of 95 fps, since you provide no thrust vs time profile.

If these assumptions are correct:

From Vf = Vo t, the time to maximum height above the 8 foot stand is 0 = 95 - 32t or t = 2.968 sec.

The height reached above the 8 foot high stand derives from h = Vot - 16t^2 or h = 95(2.968) - 16(2.968)^2 = 141 ft. or 149 feet above the ground.

The height after 4 seconds occurs at 4 - 2.968 = 1.032 sec.after reaching the maximum height.

In this 1.032 sec., the projectile falls h = 0 + 16(1.032)^2 = 17 feet.

Therefore, after 4 seconds of flight, the projectile will be 149 - 17 = 132 feet above the ground

If there are other factors that enter into the problem that you have not identified, please do so.

kaquinlan

06-08-2006, 08:18 AM

Tchrwill - the correct time is 4 minutes,not 4 seconds

I am confused - why do you have t =2.98?

TchrWill

06-08-2006, 11:16 AM

Tchrwill - the correct time is 4 minutes,not 4 seconds

I am confused - why do you have t =2.98?

A projectile leaving an 8 foot high pad with a velocity of 95 fps. will reach a zero velocity in 2.968 sec. at a height of 149 feet above the ground.

It will fall back down and hit the ground with a velocity of 97.05 fps.. in 3.05 sec.

The total flight time is therefore 5.019 sec.

4 minutes is outside the boundry of this problem as stated if a 95 fps.launch velocity is correct and the launch pad is 8 feet above the ground.

It would take a launch velocity of 7680 fps. for a projectile to reach its maximum height and zero velocity in 4 min. = 240 sec.

kaquinlan

06-09-2006, 12:18 PM

Ok you were all right! It was a typo and should be 4 seconds, not 4 minutes! So would this be correct?

A rocket takes flight from a platform that is 8ft high. It has an initial velocity of 95 ft per second. How high will the rocket be after 4 seconds of flight?

In the model, the coefficient of t^2 is one half of the acceleration due to gravity. On the surface of the earth this acceleration is approx 32 ft per second.

I know the formula is:

. . .h = -16t^2 + vt + s, where:

. . ."h" is the height (in feet) at a given time

. . ."s" is the initial height (in feet)

. . ."t" is the time in motion (in seconds)

. . ."v" is the initial velocity (in feet per second)

If the initial height is 8 ft and the initial velocity is 95 feet/sec, how high will the object be in 4 seconds?

I used:

. . .h = -16(4)^2 + 95(4) + 8

. . .h = -16(16) + 380 + 8

. . .h = -256 + 380

. . .h = 124ft?

Where does the "In the model, the coefficient of t^2 is one half of the acceleration due to gravity. On the surface of the earth this acceleration is approx 32 ft per second. "come into playor is that just an assumption already built into the equation?

The next problem is the same type but says:

On the moon the acceleration due to gravity is only 16 ft per sec. If a rocket takes off from the moon with an ititial velocity of 90 ft per sec from a height of 2ft how high will it be after 5 seconds of flight?

Do I use h = 1/2(-16t^2 )+ vt + s ?

so h= 1/2(-16(5)^2) + 90(5) + 2

h= 1/2 ((-16)25) + 450 +2

h= 1/2(-400) + 450 +2

h = -200 + 450 +2

h= 252?

TchrWill

06-09-2006, 02:07 PM

Ok you were all right! It was a typo and should be 4 seconds, not 4 minutes! So would this be correct?

A rocket takes flight from a platform that is 8ft high. It has an initial velocity of 95 ft per second. How high will the rocket be after 4 seconds of flight?

In the model, the coefficient of t^2 is one half of the acceleration due to gravity. On the surface of the earth this acceleration is approx 32 ft per second.

now the formula is:

s = -16t^2 + vt + s, where:

"h" is the height (in feet) at a given time

"s" is the initial height (in feet)

"t" is the time in motion (in seconds)

is the initial velocity (in feet per second)

If the initial height is 8 ft and the initial velocity is 95 feet/sec, how high will the object be in 4 seconds?

h = -16(4)^2 + 95(4) + 8

. . .h = -16(16) + 380 + 8

. . .h = -256 + 380

. . .h = 124ft?

The projectile reaches its maximum height in 2.968 sec. and starts to fall back to the ground. At 4 seconds, it has fallen back down 17 feet.

From Vf = Vo t, the time to maximum height above the 8 foot stand is 0 = 95 - 32t or t = 2.968 sec.

The height reached above the 8 foot high stand derives from h = Vot - 16t^2 or h = 95(2.968) - 16(2.968)^2 = 141 ft. or 149 feet above the ground.

The height after 4 seconds occurs at 4 - 2.968 = 1.032 sec.after reaching the maximum height.

In this 1.032 sec., the projectile falls h = 0 + 16(1.032)^2 = 17 feet.

Therefore, after 4 seconds of flight, the projectile will be 149 - 17 = 132 feet above the ground

kaquinlan

06-09-2006, 03:44 PM

Thanks -though this still seems confusing. So the one on the moon is incorrect as well?

kaquinlan

06-11-2006, 11:02 AM

Help! No one has responded about the one on the moon?

skeeter

06-11-2006, 05:54 PM

Help! No one has responded about the one on the moon?

The next problem is the same type but says:

On the moon the acceleration due to gravity is only 16 ft per sec. If a rocket takes off from the moon with an ititial velocity of 90 ft per sec from a height of 2ft how high will it be after 5 seconds of flight?

:roll: ... this has a mistake also.

the acceleration due to gravity on the moon's surface is not 16 ft/s<sup>2</sup>,

it is 1.6 m/s<sup>2</sup> = 5.3 ft/s<sup>2</sup>.

h(t) = 2 + 90t - 5.3t<sup>2</sup>

calculate h(5).

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